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#### Page No 163:

#### Question 1:

Suppose a charge +*Q*_{1} is given to the positive plate and a charge −*Q*_{2} to the negative plate of a capacitor. What is the "charge on the capacitor"?

#### Answer:

Given:

Charge on the positive plane = +*Q*_{1}

Charge on the negative plate = $-$*Q*_{2}

To calculate: Charge on the capacitor

Let ABCD be the Gaussian surface such that faces AD and BC lie inside plates X and Y, respectively.

Let *q* be the charge appearing on surface II. Then, the distribution of the charges on faces I, III and IV will be in accordance with the figure.

Let the area of the plates be *A* and the permittivity of the free space be${\in}_{0}$.

Now, to determine *q* in terms of *Q*_{1} and *Q*_{2}, we need to apply Gauss's law to calculate the electric field due to all four faces of the capacitor at point P. Also, we know that the electric field inside a capacitor is zero.

Electric field due to face I at point P, *E*_{1} = $\frac{{Q}_{1}-q}{2{\in}_{0}A}$

Electric field due to face II at point P, *E*_{2} = $\frac{+q}{2{\in}_{0}A}$

Electric field due to face III at point P, *E*_{3} = $\frac{-q}{2{\in}_{0}A}$

Electric field due to face IV at point P, *E*_{4} = $-\left(\frac{-{Q}_{2}+q}{2{\in}_{0}A}\right)$ (Negative sign is used as point P lies on the LHS of face IV.)

Since point P lies inside the conductor,

*E*_{1}* + **E*_{2}* + **E*_{3}* + **E*_{4}* = *0

∴ *Q*_{1} $-$ *q* + *q* $-$ *q* $-$ ($-$*Q*_{2} + *q* ) = 0

$\Rightarrow $*q* = $\frac{{Q}_{1}+{Q}_{2}}{2}$

Thus, the charge on the capacitor is $\frac{{Q}_{1}+{Q}_{2}}{2}$, which is the charge on faces II and III.

#### Page No 163:

#### Question 2:

As $C=\left(\frac{1}{V}\right)Q,$ can you say that the capacitance* C* is proportional to the charge *Q*?

#### Answer:

No. Since capacitance is a proportionality constant, it depends neither on the charge on the plates nor on the potential. It only depends upon the size and shape of the capacitor and on the dielectric used between the plates.

The formula that shows its dependence on the size and shape of the capacitor is as follows:

*C* = $\frac{{\in}_{0}A}{d}$

Here, *A* is the area of the plates of the capacitor and *d* is the distance between the plates of the capacitor.

#### Page No 163:

#### Question 3:

A hollow metal sphere and a solid metal sphere of equal radii are given equal charges. Which of the two will have higher potential?

#### Answer:

The potential of a metal sphere is directly proportional to the charge *q* given to it and inversely proportional to its radius *r.
i.e. * $V=\frac{q}{4\mathrm{\pi}{\in}_{0}r}$

Since both the spheres are conductors with the same radius and charge, the charge given to them appears on the surface evenly. Thus, the potential on the surface or within the sphere will be the same, no matter the sphere is hollow or solid.

#### Page No 163:

#### Question 4:

The plates of a parallel-plate capacitor are given equal positive charges. What will be the potential difference between the plates? What will be the charges on the facing surfaces and on the outer surfaces?

#### Answer:

It is given that the plates of the capacitor have the same charges. In other words, they are at the same potential, so the potential difference between them is zero.

Let us consider that the charge on face II is *q *so that the induced charge on face III is $-$*q** *and the distribution is according to the figure.

Now, if we consider Gaussian surface ABCD, whose faces lie inside the two plates, and calculate the field at point P due to all four surfaces, it will be

*E*_{1}$=\frac{Q-q}{2{\in}_{0}A}$

*E*_{2} = $\frac{q}{2{\in}_{0}A}$

*E*_{3} = _$\frac{q}{2{\in}_{0}A}$

*E*_{4} =_ $\frac{Q+q}{2{\in}_{0}A}$ (It is $-$ve because point P is on the left side of face IV.)

Now, as point P lies inside the conductor, the total field must be zero.

∴ *E*_{1}_{ }+ *E*_{2}_{ }+* **E*_{3}_{ }+ *E*_{4}_{ }= 0

or

*Q* $-$ *q* + *q* $-$ *q* +* Q* + *q* = 0

∴ *q* = 0

Hence, on faces II and III, the charge is equal to zero; and on faces I and IV, the charge is *Q*.

Thus, it seems that the whole charge given is moved to the outer surfaces, with zero charge on the facing surfaces.

#### Page No 163:

#### Question 5:

A capacitor has capacitance *C*. Is this information sufficient to know what maximum charge the capacitor can contain? If yes, what is this charges? If no, what other information is needed?

#### Answer:

No. This information is not sufficient. Since the charge is proportional to the potential difference across the capacitor, we need to know the potential difference applied across the capacitor.

*q *$\propto $ *V* $\Rightarrow $*q* = *CV*

Here, *q* is the charge, *V* is the potential difference applied and *C* is the proportionality constant, i.e. capacitance.

#### Page No 163:

#### Question 6:

The dielectric constant decreases if the temperature is increased. Explain this in terms of polarization of the material.

#### Answer:

The amount of polarisation can be understood as the extent of perfect alignment of the molecules of a dielectric with an external electric field. The more aligned the molecules are with the external magnetic field, the more is the polarisation and the more will be the dielectric constant.

But with increase in temperature, the thermal agitation of the molecules or the randomness in their alignment with the field increases.

Thus, we can say that increase in temperature results in reduced polarisation and reduced dielectric constant.

#### Page No 163:

#### Question 7:

When a dielectric slab is gradually inserted between the plates of an isolated parallel-plate capacitor, the energy of the system decreases. What can you conclude about the force on the slab exerted by the electric field?

#### Answer:

As the energy of the system decreases, the change in the energy is negative.

Force is defined as a negative rate of change of energy with respect to distance.

*F* = $-\frac{\partial U}{\partial x}$

So, as the energy decreases, the force due to the electric field of the capacitor increases when the dielectric is dragged into the capacitor.

#### Page No 164:

#### Question 1:

A capacitor of capacitance *C* is charged to a potential *V*. The flux of the electric field through a closed surface enclosing the capacitor is

(a) $\frac{CV}{{\epsilon}_{0}}$

(b) $\frac{2CV}{{\epsilon}_{0}}$

(c) $\frac{CV}{2{\epsilon}_{0}}$

(d) zero.

#### Answer:

(d) zero

Since the net charge enclosed by the Gaussian surface is zero, the total flux of the electric field through the closed Gaussian surface enclosing the capacitor is zero.

$\Phi =\u222fE.ds=\frac{q}{{\in}_{0}}$= 0

Here,

$\Phi $ = Electric flux

*q* = Total charge enclosed by the Gaussian surface.

#### Page No 164:

#### Question 2:

Two capacitors each having capacitance *C* and breakdown voltage *V* are joined in series. The capacitance and the breakdown voltage of the combination will be

(a) 2 *C* and 2 *V*

(b) *C*/2 and *V*/2

(c) 2 *C* and *V*/2

(d) *C*/2 and 2 *V*.

#### Answer:

(d) *C*/2 and 2*V*

Since the voltage gets added up when the capacitors are connected in series, the voltage of the combination is 2*V*.

Also, the capacitance of a series combination is given by

$\frac{1}{{C}_{\mathrm{net}}}=\frac{1}{{C}_{1}}+\frac{1}{{C}_{2}}$

Here,

*C*_{net} = Net capacitance of the combination

*C*_{1} = *C*_{2} = *C*

∴ *C*_{net}_{ }=* $\frac{C}{2}$*

#### Page No 164:

#### Question 3:

If the capacitors in the previous question are joined in parallel, the capacitance and the breakdown voltage of the combination will be

(a) 2 *C* and 2 *V*

(b) *C* and 2 *V*

(c) 2 *C* and *V*

(d) *C* and *V*.

#### Answer:

(c) 2*C* and *V*

In a parallel combination of capacitors, the potential difference across the capacitors remain the same, as the right-hand-side plates and the left-hand-side plates of both the capacitors are connected to the same terminals of the battery. Therefore, the potential remains the same, that is, *V*.

For the parallel combination of capacitors, the capacitance is given by

*C*_{eq} = *C*_{1} + *C*_{2}

Here,

*C*_{1} = *C*_{2} = *C*

∴ *C*_{eq}* *= 2*C*

#### Page No 164:

#### Question 4:

The equivalent capacitance of the combination shown in figure (31-Q1) is

(a) *C*

(b) 2 *C*

(c) *C*/2

(d) none of these.

Figure

#### Answer:

(b) 2*C*

Since the potential at point A is equal to the potential at point B, no current will flow along arm AB. Hence, the capacitor on arm AB will not contribute to the circuit. Also, because the remaining two capacitors are connected in parallel, the net capacitance of the circuit is given by

*C*_{eq}* = C* + *C* = 2*C*

#### Page No 164:

#### Question 5:

A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

(a) increase

(b) decrease

(c) remain unchanged

(d) become zero.

#### Answer:

(c) remain unchanged

The force between the plates is given by

*F* = $\frac{{q}^{2}}{2{\in}_{0}A}$

Since the capacitor is isolated, the charge on the plates remains constant.

We know that the charge is conserved in an isolated system.

Thus, the force acting between the plates remains unchanged.

#### Page No 164:

#### Question 6:

The energy density in the electric field created by a point charge falls off with the distance from the point charge as

(a) $\frac{1}{r}$

(b) $\frac{1}{{r}^{2}}$

(c) $\frac{1}{{r}^{3}}$

(d) $\frac{1}{{r}^{4}}$.

#### Answer:

(d) $\frac{1}{{r}^{4}}$

Energy density *U* is given by

*U* = $\frac{1}{2}{\in}_{0}{E}^{2}$ ...(1)

The electric field created by a point charge at a distance *r *is given by

*E* = $\frac{q}{4\mathrm{\pi}{\in}_{0}{r}^{\mathit{2}}}$

On putting the above form of *E* in eq. 1, we get

*U* = $\frac{1}{2}{\in}_{0}{\left(\frac{q}{4\mathrm{\pi}{\in}_{0}{r}^{\mathit{2}}}\right)}^{2}$

Thus, *U* is directly proportional to $\frac{1}{{r}^{4}}$.

#### Page No 164:

#### Question 7:

A parallel-plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let *Q*, and *Q* be the charges appearing on the positive and negative plates respectively.

(a) *Q*, > *Q*

(b) *Q*, = *Q*

(c) *Q*, < *Q*

(d) The information is not sufficient is decide the relation between *Q*, and *Q*.

#### Answer:

(b) ${Q}_{+}={Q}_{-}$

The charge induced on the plates of a capacitor is independent of the area of the plates.

∴ ${Q}_{+}={Q}_{-}$

#### Page No 164:

#### Question 8:

A thin metal plate *P* is inserted between the plates of a parallel-plate capacitor of capacitance *C* in such a way that its edges touch the two plates (figure 31-Q2). The capacitance now becomes

(a) *C*/2

(b) 2 *C*

(c) 0

(d) ∞.

Figure

#### Answer:

(d) $\infty $

The thin metal plate inserted between the plates of a parallel-plate capacitor of capacitance *C* connects the two plates of the capacitor; hence, the distance *d* between the plates of the capacitor reduces to zero. It can be observed that the charges on the plates begin to overlap each other via the metallic plate and hence begin to conduct continuously.

Mathematically,

*C* = $\frac{{\in}_{0}A}{d}$

In this case, *d *= 0.

∴ *C* = $\infty $

#### Page No 164:

#### Question 9:

Figure (31-Q3) shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors.

(a) *C*_{1} > *C*_{2}

(b) *C*_{1} = *C*_{2}

(c) *C*_{1} < *C*_{2}

(d) The information is not sufficient to decide the relation between *C*_{1} and *C*_{2}.

#### Answer:

(c) C_{1} < *C*_{2}

Region AB shows the potential difference across capacitor C_{1} and region CD shows the potential difference across capacitor C_{2}. Now, we can see from the graph that region AB is greater than region CD. Therefore, the potential difference across capacitor C_{1} is greater than that across capacitor C_{2}.

∵ Capacitance, *C* = $\frac{Q}{V}$

∴ C_{1} < C_{2} (*Q* remains the same in series connection.)

#### Page No 164:

#### Question 10:

Two metal plates having charges *Q*, −*Q* face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will

(a) increase

(b) decrease

(c) remain the same

(d) become zero.

#### Answer:

(a) increase

Oil between the plates of the capacitor acts as a dielectric. We know that the electric field decreases by a factor of $\frac{1}{K}$ of the original field when we insert a dielectric between the plates of a capacitor (*K* is the dielectric constant of the dielectric). So, if the oil is pumped out, the electric field between the plates will increase, as the dielectric has been removed.

#### Page No 164:

#### Question 11:

Two metal spheres of capacitance *C*_{1} and *C*_{2} carry some charges. They are put in contact and then separated. The final charges *Q*_{1} and *Q*_{2} on them will satisfy

(a) $\frac{{Q}_{1}}{{Q}_{2}}<\frac{{C}_{1}}{{C}_{2}}$

(b) $\frac{{Q}_{1}}{{Q}_{2}}=\frac{{C}_{1}}{{C}_{2}}$

(c) $\frac{{Q}_{1}}{{Q}_{2}}>\frac{{C}_{1}}{{C}_{2}}$

(d) $\frac{{Q}_{1}}{{Q}_{2}}=\frac{{C}_{2}}{{C}_{1}}$.

#### Answer:

(b) $\frac{{Q}_{1}}{{Q}_{2}}=\frac{{C}_{1}}{{C}_{2}}$

When the spheres are connected, charges flow between them until they both acquire the same common potential *V*.

The final charges on the spheres are given by

*Q*_{1}_{ }= *C*_{1}*V* and* Q*_{2} = *C*_{2}*V*

#### Page No 164:

#### Question 12:

Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are

(a) 6 µF, 18 µF

(b) 3 µF, 12 µF

(c) 2 µF, 12 µF

(d) 2 µF, 18 µF.

#### Answer:

(d) 2 µF, 18 µF

The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:

$\frac{1}{C}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$

$\Rightarrow $*C* = 2 µF

The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:

*C* = 6 + 6 + 6 = 18 µF

#### Page No 164:

#### Question 1:

The capacitance of a capacitor does not depend on

(a) the shape of the plates

(b) the size of the plates

(c) the charges on the plates

(d) the separation between the plates.

#### Answer:

(c) the charges on the plates

The capacitance of a capacitor is given by

*C* = $\frac{{\in}_{0}A}{d}$

Here, *A* is the area of the plates of the capacitor and *d* is the distance between the plates.

So, we can clearly see that the capacitance of a capacitor does depend on the size and shape of the plates and the separation between the plates; it does not depend on the charges on the plates.

#### Page No 164:

#### Question 2:

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

(a) The electric field in the capacitor

(b) The charge on the capacitor

(c) The potential difference between the plates

(d) The stored energy in the capacitor

#### Answer:

(b) The charge on the capacitor

When we insert a dielectric between the plates of a capacitor, induced charges of opposite polarity appear on the face of the dielectric. They build an electric field inside the dielectric, directed opposite to the original field of the capacitor.

Thus, the net effect is a reduced electric field.

Also, as the potential is proportional to the field, the potential decreases and so does the stored energy *U*, which is given by

*U* = $\frac{qV}{2}$

Thus, only the charge on the capacitor remains unchanged, as the charge is conserved in an isolated system.

#### Page No 165:

#### Question 3:

A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is *Q* and the magnitude of the induced charge on each surface of the dielectric is *Q*'.

(a) *Q*' may be large than *Q*.

(b) *Q*' must be larger than *Q*.

(c) *Q*' must be equal to *Q*.

(d) *Q*' must be smaller than *Q*.

#### Answer:

(d) *Q*' must be smaller than *Q*.

The relation between the induced charge *Q*' and the charge on the capacitor* Q* is given by

*Q*' = $Q\left(1-\frac{1}{K}\right)$

Here, *K* is the dielectric constant that is always greater than or equal to 1.

So, we can see that for* K* > 1, *Q*' will always be less than *Q*.

#### Page No 165:

#### Question 4:

Each plate of a parallel plate capacitor has a charge* q* on it. The capacitor is now connected to a batter. Now,

(a) the facing surfaces of the capacitor have equal and opposite charges

(b) the two plates of the capacitor have equal and opposite charges

(c) the battery supplies equal and opposite charges to the two plates

(d) the outer surfaces of the plates have equal charges

#### Answer:

(a) the facing surfaces of the capacitor have equal and opposite charges

(b) the two plates of the capacitor have equal and opposite charges

(d) the outer surfaces of the plates have equal charges

In H.C Verma the answer is (a), (c) ,(d). But according to us the answer should be (a), (b), (d) all these options are the properties of a capacitor and the option (c) is incorrect because the battery is a source of energy not charge. Moreover if a capacitor plates have equal charge on outside and equal charge on inside then one can think that the charge on the plates must be also equal so option (b) cant be incorrect.

#### Page No 165:

#### Question 5:

The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change?

(a) Charge on the capacitor

(b) Potential difference across the capacitor

(c) Energy of the capacitor

(d) Energy density between the plates

#### Answer:

(b) Potential difference across the capacitor

(c) Energy of the capacitor.

Because the charge always remains conserved in an isolated system, it will remain the same.

Now,

*V*=$\frac{Qd}{{\in}_{0}A}$

Here, *Q*, *A* and *d *are the charge, area and distance between the plates, respectively.

Thus, as *d* increases, *V* increases.

Energy is given by

*E* = $\frac{qV}{2}$

So, it will also increase.

Energy density* u*, that is, energy stored per unit volume in the electric field is given by

*u* = $\frac{1}{2}{\in}_{0}{E}^{2}$

So, *u *will remain constant with increase in distance between the plates.

#### Page No 165:

#### Question 6:

A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor.

(a) The battery will supply more charge.

(b) The capacitance will increase.

(c) The potential difference between the plates will increase.

(d) Equal and opposite charges will appear on the two faces of the metal plate.

#### Answer:

(d) Equal and opposite charges will appear on the two faces of the metal plate.

The capacitance of the capacitor in which a dielectric slab of dielectric constant *K*, area *A* and thickness* t* is inserted between the plates of the capacitor of area *A* and separated by a distance *d* is given by

*C* = $\frac{{\in}_{0}A}{\left(d-t\right)+\left(t/K\right)}$

Since it is given that the thickness of the sheet is negligible, the above formula reduces to *C* = $\frac{{\in}_{0}A}{d}$. In other words, there will not be any change in the electric field, potential or charge.

Only, equal and opposite charges will appear on the two faces of the metal plate because of induction due to the presence of the charges on the plates of the capacitor.

#### Page No 165:

#### Question 7:

Following operations can be performed on a capacitor:

*X* − connect the capacitor to a battery of emf ε.

*Y* − disconnect the battery.

*Z* − reconnect the battery with polarity reversed.

*W* − insert a dielectric slab in the capacitor.

(a) In *XYZ* (perform *X*, then *Y*, then *Z*) the stored electric energy remains unchanged and no thermal energy is developed.

(b) The charge appearing on the capacitor is greater after the action *XWY *than after the action *XYZ.*

(c) The electric energy stored in the capacitor is greater after the action *WXY* than after the action *XYW*.

(d) The electric field in the capacitor after the action *XW* is the same as that after *WX*.

#### Answer:

(b) The charge appearing on the capacitor is greater after the action *XWY *than after the action *XYZ.*

(c) The electric energy stored in the capacitor is greater after the action *WXY* than after the action *XYW*.

(d) The electric field in the capacitor after the action *XW* is the same as that after *WX*.

Justification of option (b)

If the potential is held constant, that is, the battery remains attached to the circuit, then the charge on the capacitor increases by a factor of *K *on inserting a dielectric of a dielectric constant *K* between the plates of the capacitor.

Mathematically,

*q = Kq _{0 }*

Here,

*q*and

_{0}*q*are the charges without dielectric and with dielectric, respectively.

The amount of charge stored does not depend upon the polarity of the plates.

Thus, the charge appearing on the capacitor is greater after the action

*XWY*than after the action

*XYZ.*

Justification of option (c)

Since the battery is disconnected before inserting a dielectric, the amount of charge remains constant, that is,

*q = q*, because after the battery is disconnected, the capacitor gets no source to store charge from. In other words, the capacitor is now an isolated system where the amount of charge is conserved and so is the energy

_{0}*U*as

*$\frac{1}{2}$q $\epsilon $*. Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor. So, the energy stored in the capacitor will also not change after the action

*XYW*.

However, during the action

*WXY,*the amount of charge that will get stored in the capacitor will get increased by a factor of

*K,*as the battery is disconnected after inserting a dielectric between the plates of the capacitor and the energy stored will also get multiplied by a factor of

*K.*

Thus, the electric energy stored in the capacitor is greater after the action

*WXY*than after the action

*XYW.*

Justification of option (d)

The electric field between the plates

*E*depends on the potential across the capacitor and the distance

*d*between the plates of the capacitor.

Mathematically,

*E = $\frac{\epsilon}{d}$*

In either case, that is, during actions

*XW*and

*WX,*the potential remains the same, that is, $\epsilon $. Thus, the electric field

*E*remains the same.

Denial of option (a)

During the action

*XYZ*, the battery has to do extra work equivalent to $\frac{1}{2}$

*CV*to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be $\frac{1}{2}$

^{2}*CV*

^{2}^{ }+ $\frac{1}{2}$

*CV*. This extra work done will be dissipated as heat energy. Thus, thermal energy is developed. However, the stored electric energy remains unchanged, that is, $\frac{1}{2}$

^{2}*CV*

^{2}.#### Page No 165:

#### Question 1:

When 1⋅0 × 10^{12} electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Calculate the capacitance of the two-conductor system.

#### Answer:

As 1.0 × 10^{12 }electrons are transferred from one conductor to another, the conductor to which the electrons are transferred becomes negatively charged and the other conductor becomes positively charged.

Now,

Magnitude of the net charge on each conductor, *Q* = (1.0 × 10^{12}) × (1.6 × 10^{$-$19}) C = 1.6 × 10^{$-$7} C

Magnitude of the potential difference between the conductors, *V* = 10 V

The capacitance *C* is the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between the conductors.

$C=\frac{Q}{V}$

$\Rightarrow C=\frac{1.6\times {10}^{-7}}{10}=1.6\times {10}^{-8}\mathrm{F}$

Hence, the value of the capacitance of the given two conductor systems is 1.6 × 10^{$-$8} F.

#### Page No 165:

#### Question 2:

The plates of a parallel-plate capacitor are made of circular discs of radii 5⋅0 cm each. If the separation between the plates is 1⋅0 mm, what is the capacitance?

#### Answer:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{{\in}_{0}A}{d}$

Here,

*A* = Area of the plate

*d* = Distance between the parallel plates

Given:

$\mathrm{A}=\mathrm{\pi}{r}^{2}=\mathrm{\pi}\times {\left(5\times {10}^{-2}\right)}^{2}\phantom{\rule{0ex}{0ex}}d=1.0\times {10}^{-3}\mathrm{m}\phantom{\rule{0ex}{0ex}}{\in}_{0}=8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\phantom{\rule{0ex}{0ex}}\therefore C=\frac{{\in}_{0}\mathit{}A}{d}\phantom{\rule{0ex}{0ex}}=\frac{8.85\times {10}^{-12}\times 3.14\times 25\times {10}^{-4}}{{10}^{-3}}\phantom{\rule{0ex}{0ex}}=6.95\times {10}^{-5}\mathrm{\mu F}$

#### Page No 165:

#### Question 3:

Suppose, one wishes to construct a 1⋅0 farad capacitor using circular discs. If the separation between the discs be kept at 1⋅0 mm, what would be the radius of the discs?

#### Answer:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{{\in}_{0}A}{d}$

Here,

*A* = Area of the plate

*d* = Distance between the parallel plates

Now,

Let the radius of the disc be *r.*

$\therefore C=\frac{{\in}_{0}A}{d}=\frac{{\in}_{0}\left(\mathrm{\pi}{r}^{2}\right)}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{\frac{Cd}{{\in}_{0}\mathrm{\pi}}}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{\frac{1\times (1\times {10}^{-3})}{8.85\times {10}^{-12}\times 3.14}}=\sqrt{35.98\times {10}^{6}}\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow r\approx \sqrt{36\times {10}^{6}}\mathrm{m}=6\times {10}^{3}\mathrm{m}=6\mathrm{km}\phantom{\rule{0ex}{0ex}}$

Thus, the radius of the plates of the capacitor for the given configuration is 6 km.

#### Page No 165:

#### Question 4:

A parallel-plate capacitor having plate area 25 cm^{2}^{ }and separation 1⋅00 mm is connected to a battery of 6⋅0 V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?

#### Answer:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{{\in}_{0}A}{d}$

Here,

*A* = Area of the plate

*d* = Distance between the parallel plates

Given:

*A* = 25 cm^{2} = 25 × 10^{$-$4} m^{2}

*d* = 1.00 mm = 1 × 10^{$-$3} m

Now,

$C=\frac{{\in}_{0}\mathrm{A}}{d}=\frac{8.85\times {10}^{-12}\times 25\times {10}^{-4}}{1\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=2.21\times {10}^{-11}\mathrm{F}$

When the battery of voltage 6 V is connected to the capacitor, the charge (*Q*) that flows from the battery is equal to the amount of the charge that the given capacitor can hold.

$\Rightarrow $*Q* = *CV*

$\Rightarrow Q=2.21\times {10}^{-11}\times 6.0\phantom{\rule{0ex}{0ex}}=1.33\times {10}^{-10}\mathrm{C}$

The work done by the battery in charging the capacitor is calculated by taking the product of the magnitude of the charge transferred and the voltage of the battery.

Thus, we get

$W=QV\phantom{\rule{0ex}{0ex}}=1.33\times {10}^{-10}\times 6.0\phantom{\rule{0ex}{0ex}}=8.0\times {10}^{-10}\mathrm{J}$

Thus, the charge flown through the battery is 1.33 × 10^{$-$10} C and the work done by the battery is 8.0 × 10^{$-$10 }J.

#### Page No 165:

#### Question 5:

A parallel-plate capacitor has plate area 25⋅0 cm^{2} and a separation of 2⋅00 mm between the plates. The capacitor is connected to a battery of 12⋅0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1⋅00 mm. Find the extra charge given by the battery to the positive plate.

#### Answer:

Given:

Area of the plate, *A* = 25 cm^{2} = 25 × 10^{$-$4} m^{2}

Separation between the plates, *d* = 2 mm = 2 × 10^{$-$3} m

Potential difference between the plates, *V* = 12 V

The capacitance of the given capacitor is given by

$C=\frac{{\in}_{0}A}{d}$

$=\frac{(8.85\times {10}^{-12})\times (25\times {10}^{-4})}{(2\times {10}^{-3})}\phantom{\rule{0ex}{0ex}}=11.06\times {10}^{-12}\mathrm{F}$

(a) Charge on the capacitor is given by

*Q = CV*

$=11.06\times {10}^{-12}\times 12\phantom{\rule{0ex}{0ex}}=1.33\times {10}^{-10}\mathrm{C}$

(b) When the separation between the plates is decreased to 1 mm, the capacitance C' can be calculated as:

$C\text{'}=\frac{{\in}_{0}\mathrm{A}}{d}\phantom{\rule{0ex}{0ex}}=\frac{(8.85\times {10}^{-12})\times (25\times {10}^{-4})}{1\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=22.12\times {10}^{-12}\mathrm{F}\phantom{\rule{0ex}{0ex}}$

Charge on the capacitor is given by

*Q*' = *C'V*

$=22.12\times {10}^{-12}\times 12\phantom{\rule{0ex}{0ex}}=2.65\times {10}^{-10}\mathrm{C}\phantom{\rule{0ex}{0ex}}\mathrm{Extra}\mathrm{charge}\phantom{\rule{0ex}{0ex}}=\left(2.65\times {10}^{-10}-1.32\times {10}^{-10}\right)\mathrm{C}\phantom{\rule{0ex}{0ex}}=1.33\times {10}^{-10}\mathrm{C}$

#### Page No 165:

#### Question 6:

Find the charges on the three capacitors connected to a battery as shown in figure (31-E1) Take ${C}_{1}=2\xb70\mu \mathrm{F},{C}_{2}=4\xb70\mu \mathrm{F},{C}_{3}=6\xb70\mu \mathrm{F}\mathrm{and}V=12\mathrm{volts}.$

Figure

#### Answer:

The capacitances of three capacitors are *C*_{1}= 2 μF, *C*_{2}= 4 μF and *C*_{3}= 6 μF and the voltage of the battery (*V*) is 12 V.

As the capacitors are connected in parallel, the equivalent capacitance is given by

*C*_{eq}* = **C*_{1}* + **C*_{2}* + **C*_{3}

= (2 + 4 + 6) μF = 12 μF = 12 × 10^{$-$6 }F

Due to parallel connection, the potential difference across each capacitor is the same and is equal to 12 V.

Therefore, the charge on each capacitor can be calculated as follows:

The charge on the capacitor of capacitance *C*_{1}= 2 μF is given by

*Q*_{1}_{ }= *C*_{1}*V* = (2 × 10^{$-$6}) × 12 C = 24 × 10^{$-$6} C = 24 μC

Similarly, the charges on the other two capacitors are given by

*Q*_{2 }= *C*_{2}*V* = (4 × 10^{$-$6}) × 12 C = 48 × 10^{$-$6} C = 48 μC

and

*Q*_{3}_{ }= *C*_{3}*V* = (6 × 10^{$-$6}) × 12 C = 72 × 10^{$-$6} C = 72 μC

#### Page No 165:

#### Question 7:

Three capacitors having capacitances 20 µF, 30 µF and 40 µF are connected in series with a 12 V battery. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors?

#### Answer:

When the capacitors are connected in series, the equivalent capacitance is given by

$\frac{1}{{c}_{eq}}=\frac{1}{{c}_{1}}+\frac{1}{{c}_{2}}+\frac{1}{{c}_{3}}\phantom{\rule{0ex}{0ex}}\frac{1}{{c}_{eq}}=\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{40}\right)\times \frac{1}{{10}^{-6}}\phantom{\rule{0ex}{0ex}}\Rightarrow {c}_{eq}=9.23\mathrm{\mu F}$

Because the capacitors are connected in series, the same charge will go to each of them and it is equal to the total charge given by the battery.

Now,

Let the charge at each capacitor be *q.*

$\therefore q=CV=(9.23\times {10}^{-6})\times 12\phantom{\rule{0ex}{0ex}}q=110.76\mathrm{\mu C}$

The work done by the battery (*W*) is given by

$W\mathit{=}qV\phantom{\rule{0ex}{0ex}}\Rightarrow W=12\times 110.76\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow W=1.33\times {10}^{-3}\mathrm{J}$

#### Page No 165:

#### Question 8:

Find the charge appearing on each of the three capacitors shown in figure (31-E2).

Figure

#### Answer:

Let us first find the equivalent capacitance. It can be observed from the circuit diagram that capacitors B and C are in parallel and are in series with capacitor A.

The equivalent capacitance can be calculated as follows:

$\frac{1}{{C}_{\mathrm{eq}}}=\frac{1}{{C}_{\mathrm{A}}}+\frac{1}{{C}_{\mathrm{B}}+{C}_{\mathrm{C}}}\phantom{\rule{0ex}{0ex}}\frac{1}{{C}_{\mathrm{eq}}}=\frac{1}{8}+\frac{1}{4+4}=\frac{1}{8}+\frac{1}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{C}_{\mathrm{eq}}}=\frac{2}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{eq}}=4\mathrm{\mu}F$

Capacitors B and C are parallel and are in series with capacitor A. The equivalent capacitance of capacitors B and C is given by

(4 + 4) μF = 8 μF

It is the same as the capacitance of capacitor A. Therefore, equal potential difference will be there on capacitor A and the system of capacitors B and C.

Now,

Potential difference across capacitor A = 6 V

Thus,

Charge on capacitor A = (8 µF) × (6 V) = 48 µC

And,

Potential difference across capacitors B and C = 6 V

Thus,

Charge on capacitor B = (4 µF) × (6 V) = 24 µF

Charge on capacitor C = (4 µF) × (6 V) = 24 µF

#### Page No 165:

#### Question 9:

Take ${C}_{1}=4\xb70\mu \mathrm{F}\mathrm{and}{C}_{2}=6\xb70\mu \mathrm{F}$ in figure (31-E3). Calculate the equivalent capacitance of the combination between the points indicated.

Figure

#### Answer:

(a)

For the combination of capacitors given in figure (a), the pairs of capacitors *C*_{1}_{ }and *C*_{2}_{ }are in parallel.

The equivalent capacitance of each parallel combination of capacitors is given by

*C*_{1}_{ }+ *C*_{2} = 4 + 6 = 10 μF

The equivalent circuit can be drawn as:

The equivalent capacitance for the above series circuit is given by

$\frac{1}{{C}_{\mathrm{eq}}}=\frac{1}{{C}_{1}+{C}_{2}}+\frac{1}{{C}_{1}+{C}_{2}}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{eq}}=5\mathrm{\mu F}$

(b)

For the combination of capacitors given in figure (b), the pairs of capacitors *C*_{1}_{ }and *C*_{2}_{ }are in parallel.

The equivalent capacitance of each parallel combination of capacitors is given by

*C*_{1}_{ }+ *C*_{2} = 4 + 6 = 10 μF

The equivalent circuit can be drawn as:

In the above circuit, it can be seen that *C*_{A }and *C*_{B} are in series and are in parallel to the series combination of *C*_{C }and *C*_{B}.

The equivalent capacitance for the series combination of *C*_{A }and *C*_{B }is given by

$\frac{1}{{C}_{\mathrm{eq}}}=\frac{1}{{C}_{\mathrm{A}}}+\frac{1}{{C}_{\mathrm{B}}}=\frac{1}{{C}_{1}+{C}_{2}}+\frac{1}{{C}_{1}+{C}_{2}}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{eq}}=5\mathrm{\mu F}$

Similarly, the equivalent capacitance of the series combination of *C*_{C }and *C*_{D }is 5 μF.

∴ Net equivalent capacitance = 5 + 5 = 10 μF

#### Page No 166:

#### Question 10:

Find the charge supplied by the battery in the arrangement shown in figure (31-E4).

Figure

#### Answer:

The equivalent circuit for the given case can be drawn as:

It can be observed that capacitors* **C*_{1}_{ }and *C*_{2} are in parallel.

Therefore, the equivalent capacitance is given by

*C*_{eq} = *C*_{1}_{ }+ *C*_{2}

= 5 + 6 = 11 μF

Now,

Charge supplied by the battery = (*C*_{eq})(V) = (11 μF) × (10 V) = 110 μC

#### Page No 166:

#### Question 11:

The outer cylinders of two cylindrical capacitors of capacitance 2⋅2 µF each, are kept in contact and the inner cylinders are connected through a wire. A battery of emf 10 V is connected as shown in figure (31-E5). Find the total charge supplied by the battery to the inner cylinders.

Figure

#### Answer:

It is given that the outer cylinders are kept in contact and the inner cylinders are connected through a wire. Thus, the equivalent circuit can be drawn as:

The left plate of the capacitors (A and B) shown in the figure represents the inner plates of the cylindrical capacitors.

As the capacitors are connected in parallel, the potential difference across them is the same.

∴ Magnitude of the charge on each capacitor = *CV* = (2.2 μF) × (10V) = 22 µC

As plates A and B are connected to the positive terminal of the battery, the charge on each of them is +22 µC.

∴ Net charge on the inner plates = 22 µC + 22 µC = +44 µC

#### Page No 166:

#### Question 12:

Two conducting spheres of radii *R*_{1} and *R*_{2} are kept widely separated from each other. What are their individual capacitances? If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series−parallel connections.

#### Answer:

We need to calculate the capacitance of an isolated charged sphere.

Let us assume that the charge on the sphere is *Q* and its radius is *R*.

Capacitance of the charged sphere can be found by imagining a concentric sphere of infinite radius consisting of −Q charge.

Potential difference between the spheres = $\frac{1}{4\pi {\in}_{0}}\frac{Q}{R}$ − 0 = $\frac{1}{4\pi {\in}_{0}}\frac{Q}{R}$

Capacitance is the ratio of the magnitude of the charge on each conductor to the potential difference between them.

$C=\frac{Q}{{\displaystyle \frac{1}{4\pi {\in}_{0}}\frac{Q}{R}}}=4\pi {\in}_{0}R$

Therefore, the capacitances of spheres of radii *R*_{1} and *R*_{2} are *C*_{1} and *C*_{2}, respectively. They are given by

${C}_{1}=4\pi {\in}_{0}{R}_{1}\phantom{\rule{0ex}{0ex}}{C}_{2}=4\mathrm{\pi}{\in}_{0}{R}_{2}$

If the spheres are connected by a metal wire, the charge will flow from one sphere to another till their potentials become the same.

As there potentials become the same, the potential difference between the conductors for both the capacitors also becomes the same. Thus, it can be concluded that the capacitors are connected in parallel.

Thus, the capacitance of the combination is given by

${C}_{\mathrm{eq}}$ = *C*_{1} + *C*_{2}

$=4\mathrm{\pi}{\in}_{0}\left({R}_{1}+{R}_{2}\right)$

#### Page No 166:

#### Question 13:

Each of the capacitors shown in figure (31-E6) has a capacitance of 2 µF. find the equivalent capacitance of the assembly between the points *A* and *B*. Suppose, a battery of emf 60 volts is connected between *A* and *B*. Find the potential difference appearing on the individual capacitors.

Figure

#### Answer:

There are three rows of capacitors connected in parallel in the given system. In each row, three capacitors of capacitance 2 μF are connected in series.

For each row, the equivalent capacitance is given by

$\frac{1}{{C}_{\mathrm{r}}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{r}}=\frac{2}{3}\mathrm{\mu F}$

As three rows are connected in parallel, their equivalent capacitance is given by

*C*_{eq}_{ }= *C*_{r} + *C*_{r} + *C*_{r} = $\frac{2}{3}+\frac{2}{3}+\frac{2}{3}$ = 2 μF

The voltage across each row is the same and is equal to 60 V.

As all capacitors have the same capacitance in each row, the potential difference across their plates is the same.

∴ Potential difference across each capacitor = 20 V

#### Page No 166:

#### Question 14:

It is required to construct a 10 µF capacitor which can be connected across a 200 V battery. Capacitors of capacitance 10 µF are available but they can withstand only 50 V. Design a combination which can yield the desired result.

#### Answer:

Let the number of capacitors in series (connected in a row) be *x*.

The maximum voltage that the capacitors can withstand is 50 V.

The voltage across each row should be equal to 200 V.

Therefore,

*x* × 50 = 200

Thus,

*x* = 4 capacitors

Now,

Let there be *y* such rows.

So, the equivalent capacitance of the combination will be *xy.*

⇒* xy *= 10

⇒ *y* = 10 *x* = 4 capacitors

Thus, to yield the required result, the combination of 4 rows, each of 4 capacitors having capacitance 10 µF and breakdown voltage 50 V, is required.

#### Page No 166:

#### Question 15:

Take the potential of the point *B* in figure (31-E7) to be zero. (a) Find the potentials at the points *C* and *D*. (b) If a capacitor is connected between *C* and *D*, what charge will appear on this capacitor?

Figure

#### Answer:

(a)

The capacitance of the two rows connected in parallel is given by

${C}_{1}=\frac{4\times 8}{4+8}=\frac{8}{3}\mathrm{\mu F}$ and ${C}_{2}=\frac{3\times 6}{3+6}=\frac{18}{9}\mathrm{\mu F}=2\mathrm{\mu F}$

As the two rows are in parallel, the potential difference across each row is the same and is equal to 50 V.

The charge on the branch ACB with capacitance $\frac{8}{3}\mathrm{\mu F}$ is given by

*Q* = $\left(\frac{8}{3}\mathrm{\mu F}\right)$×(50 V) = $\frac{400}{3}\mathrm{\mu C}$

The charge on the branch ADB with capacitance $2\mathrm{\mu F}$ is given by

$Q=C\times V\phantom{\rule{0ex}{0ex}}Q=2\mathrm{\mu F}\times 50=100\mathrm{\mu C}$

The potential at point D is given by

${V}_{\mathrm{D}}=\frac{q}{{\mathrm{C}}_{1}}=\frac{100\mathrm{\mu C}}{6\mathrm{\mu F}}\phantom{\rule{0ex}{0ex}}{V}_{\mathrm{D}}=\frac{50}{3}\mathrm{V}$

Similarly, the potential at point C is given by

${V}_{\mathrm{C}}=\frac{50}{3}\mathrm{V}$

(b) As the potential difference between points C and D is zero, the bridge remains balanced and no charge flows from C to D. If a capacitor is connected between points C and D, then the change on the capacitor will be zero.

#### Page No 166:

#### Question 16:

Find the equivalent capacitance of the system shown in figure (31-E8) between the points *a* and *b*.

Figure

#### Answer:

Capacitors *C*_{1}_{ }and C_{2} above and below *ab *are connected in series and in parallel with *C*_{3}.

Thus, the equivalent capacitor between *a* and *b* is given by

${C}_{\mathrm{eq}}=\left(\frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}+{C}_{3}+\frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{eq}}={C}_{3}+\frac{2{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}$

#### Page No 166:

#### Question 17:

A capacitor is made of a flat plate of area *A* and a second plate having a stair-like structure as shown in figure (31-E9). The width of each stair is *a* and the height is *b*. Find the capacitance of the assembly.

Figure

#### Answer:

The total area of the flat plate is *A*. The width of each stair is the same. Therefore, the area of the surface of each stair facing the flat plate is the same, that is, $\frac{A}{3}$.

From the figure, it can be observed that the capacitor assembly is made up of three capacitors. The three capacitors are connected in parallel.

For capacitor *C*_{1}, the area of the plates is $\frac{A}{3}$ and the separation between the plates is *d*.

For capacitor *C*_{2}, the area of the plates is $\frac{A}{3}$ and the separation between the plates is (*d + b*).

For capacitor *C*_{3}, the area of the plates is $\frac{A}{3}$ and the separation between the plates is (*d *+ 2*b*).

Therefore,

${C}_{1}=\frac{{\in}_{0}A}{3d}\phantom{\rule{0ex}{0ex}}{C}_{2}=\frac{{\in}_{0}A}{3\left(d+b\right)}\phantom{\rule{0ex}{0ex}}{C}_{3}=\frac{{\in}_{0}\mathit{}A}{3\left(d+2b\right)}$

As the three capacitors are in parallel combination,

$C={C}_{1}+{C}_{2}+{C}_{3}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\in}_{0}A}{3d}+\frac{{\in}_{0}A}{3\left(d+b\right)}+\frac{{\in}_{0}A}{3\left(d+2b\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\in}_{0}A}{3}\frac{\left(3{d}^{2}+6bd+2{b}^{2}\right)}{d\left(d+b\right)\left(d+2b\right)}$

#### Page No 166:

#### Question 18:

A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10 cm and of radii 2 mm and 4 mm. (a) Calculate the capacitance. (b) Another capacitor of the same length is constructed with cylinders of radii 4 mm and 8 mm. Calculate the capacitance.

#### Answer:

(a) The capacitance of a cylindrical capacitor is given by

$C=\frac{2\mathrm{\pi}{\in}_{0}l}{\mathrm{ln}\left({\displaystyle \frac{{R}_{2}}{{R}_{1}}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{2\times 3.14\times 8.85\times {10}^{-12}\times 0.10}{\mathrm{ln}2}\phantom{\rule{0ex}{0ex}}\Rightarrow C=8\mathrm{pF}\left(\because \mathrm{l}\mathrm{n}2=0.693\right)$

(b) When a capacitor of the same height with cylinders of radii 4 mm and 8 mm is taken, its capacitance comes to 8 pF, which is the same as above because the ratio of the radii is the same.

#### Page No 166:

#### Question 19:

A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20 pF. What will be the new potential difference across the 100 pF capacitor?

#### Answer:

Given:

${C}_{1}=100\mathrm{pF}\phantom{\rule{0ex}{0ex}}V=24\mathrm{V}$

Charge on the given capacitor, $q={C}_{1}V=24\times 100\mathrm{pC}$

Capacitance of the uncharged capacitor,${C}_{2}=20\mathrm{pF}$

When the charged capacitor is connected with the uncharged capacitor, the net charge on the system of the capacitors becomes

${q}_{1}+{q}_{2}=24\times 100\mathrm{qC}...\left(\mathrm{i}\right)$

The potential difference across the plates of the capacitors will be the same.

Thus,

$\frac{{q}_{1}}{{C}_{1}}=\frac{{q}_{2}}{{C}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{q}_{1}}{100}=\frac{{q}_{2}}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow {q}_{1}=5{q}_{2}...\left(\mathrm{ii}\right)$

From eqs. (i) and (ii), we get

${q}_{1}+\frac{{q}_{1}}{5}=24\times 100\mathrm{pC}\phantom{\rule{0ex}{0ex}}\Rightarrow 6{q}_{1}=5\times 24\times 100\mathrm{pC}\phantom{\rule{0ex}{0ex}}\Rightarrow {q}_{1}=\frac{5\times 24\times 100}{6}\mathrm{pC}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{V}_{1}=\frac{{q}_{1}}{{C}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{5\times 24\times 100\mathrm{pC}}{6\times 100\mathrm{pF}}=20\mathrm{V}$

#### Page No 166:

#### Question 20:

Each capacitor shown in figure (31-E10) has a capacitance of 5⋅0 µF. The emf of the battery is 50 V. How much charge will flow through *AB* if the switch *S* is closed?

Figure

#### Answer:

Initially, when the switch S is open, the equivalent capacitance is given by

${C}_{\mathrm{eq}}=\frac{2C\times C}{3\mathrm{C}}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{eq}}=\frac{2}{3}C=\frac{2}{3}\times 5.0\mathrm{\mu F}$

The charge supplied by the battery is given by

$Q={C}_{\mathrm{eq}}\times V\phantom{\rule{0ex}{0ex}}\Rightarrow Q=\frac{2}{3}\times (5.0\mathrm{\mu F})\times (50\mathrm{V})\phantom{\rule{0ex}{0ex}}\Rightarrow Q=\frac{500}{3}\mathrm{\mu C}$

When the switch S is closed, no charge goes to the capacitor connected in parallel with the switch.

Thus, the equivalent capacitance is given by

${C}_{\mathrm{eq}}=2C=2\times 5.0=10\mathrm{\mu F}$

The charge supplied by the battery is given by

$Q=10\mathrm{\mu F}\times 50=500\mathrm{\mu C}$

The initial charge stored in the shorted capacitor starts discharging."?

Hence, the charge that flows from A to B is given by

${Q}_{\mathrm{net}}=500\mathrm{\mu C}-\frac{500}{3}\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}\Rightarrow {Q}_{\mathrm{net}}=3.3\times {10}^{-4}\mathrm{C}$

#### Page No 166:

#### Question 21:

The particle *P* shown in figure (31-E11) has a mass of 10 mg and a charge of −0⋅01 µC. Each plate has a surface area 100 cm^{2} on one side. What potential difference *V* should be applied to the combination to hold the particle *P* in equilibrium?

Figure

#### Answer:

The particle is balanced when the electrical force on it is balanced by its weight.

Thus,

$mg=qE\phantom{\rule{0ex}{0ex}}mg=q\times \frac{V\text{'}}{d}...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}$

Here,

*d* = Separation between the plates of the capacitor

*V'* = Potential difference across the capacitor containing the particle

We know that the capacitance of a capacitor is given by

$C=\frac{{\in}_{0}A}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow d=\frac{{\in}_{0}A}{C}$

Thus, eq. (*i*) becomes

$mg=q\times V\text{'}\times \frac{\mathrm{C}}{{\in}_{0}A}\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=\frac{mg{\in}_{0}A}{q\times C}\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=\frac{{10}^{-6}\times 9.8\times (8.85\times {10}^{-12})\times (100\times {10}^{-4})}{(0.01\times {10}^{-6})\times (0.04\times {10}^{-6})}\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=21.68\mathrm{mV}$

Since the values of both the capacitors are the same,

*V* = 2*V*' = 2 $\times $ 21.86 $\approx $ 43 mV

#### Page No 166:

#### Question 22:

Both the capacitors shown in figure (31-E12) are made of square plates of edge *a*. The separations between the plates of the capacitors are *d*_{1} and *d*_{2} as shown in the figure. *A* potential difference *V* is applied between the points *a* and *b*. An electron is projected between the plates of the upper capacitor along the central line. With what minimum speed should the electron be projected so that it does not collide with any plate? Consider only the electric forces.

Figure

#### Answer:

Let:

Velocity of the electron = *u*

Mass of the electron = *m*

Now,

Horizontal distance, *x* = *u* × *t*

$\Rightarrow t=\frac{x}{v}$ ...(i)

Let the electric field inside the capacitor be *E*

∴ Acceleration of the electron =$\frac{qE}{m}$

Vertical distance, $y=\frac{1}{2}\frac{qE}{m}{t}^{2}=\frac{1}{2}\frac{qE}{m}{\left(\frac{x}{u}\right)}^{2}\left(\because t=\frac{x}{u}\right)$

$y=\frac{{d}_{1}}{2}\mathrm{and}x=a\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{d}_{1}}{2}=\frac{1}{2}\frac{qE}{m}\xb7{\left(\frac{a}{u}\right)}^{2}...\left(\mathrm{ii}\right)$

Capacitance of the two capacitors:

*C*_{1} = $\frac{{\in}_{0}{a}^{2}}{{d}_{1}}$ and *C*_{2} = $\frac{{\in}_{0}{a}^{2}}{{d}_{2}}$

It is given that the capacitors are connected in series.

Thus, the equivalent capacitance is given by

${C}_{\mathrm{eq}}=\frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}$

${C}_{\mathrm{eq}}=\frac{{\displaystyle \frac{{\in}_{0}{a}^{2}}{{d}_{1}}}\times {\displaystyle \frac{{\in}_{0}{a}^{2}}{{d}_{2}}}}{{\displaystyle \frac{{\in}_{0}{a}^{2}}{{d}_{1}}}+{\displaystyle \frac{{\in}_{0}{a}^{2}}{{d}_{2}}}}=\frac{{\in}_{0}{a}^{2}}{\left({d}_{1}+{d}_{2}\right)}$

Total charge on the system of capacitors, *Q* = *C*_{eq}*V* = $\frac{{\in}_{0}{a}^{2}}{\left({d}_{1}+{d}_{2}\right)}V$

As the capacitors are in series, charge on both of them is the same.

The potential difference across the capacitor containing the electron is given by

$V=\frac{Q}{{C}_{1}}=\frac{{\in}_{0}{a}^{2}V}{{C}_{1}\left({d}_{1}+{d}_{2}\right)}=\frac{{\in}_{0}{a}^{2}V}{\left({\displaystyle \frac{{\in}_{0}{a}^{2}}{{d}_{1}}}\right)\left({d}_{1}+{d}_{2}\right)}=\frac{V{d}_{1}}{{d}_{1}+{d}_{2}}$

The magnitude of the electric field inside the capacitor is given by

$E=\frac{V}{{d}_{1}}=\frac{V}{{d}_{1}+{d}_{2}}$

The charge on electron *q* is represented by *e.*

On putting the values of *q* and *E* in (ii), we get

$\Rightarrow \frac{{d}_{1}}{2}=\frac{1}{2}\frac{qV}{m({d}_{1}+{d}_{2})}\xb7{\left(\frac{a}{u}\right)}^{2}...\left(\mathrm{iii}\right)$

The minimum velocity of the electron is given by

$u={\left(\frac{Ve{a}^{2}}{m{d}_{1}\left({d}_{1}+{d}_{2}\right)}\right)}^{1/2}$

#### Page No 167:

#### Question 23:

The plates of a capacitor are 2⋅00 cm apart. An electron-proton pair is released somewhere in the gap between the plates and it is found that the proton reaches the negative plate at the same time as the electron reaches the positive plate. At what distance from the negative plate was the pair released?

#### Answer:

Let the electric field inside the capacitor be *E*.

Now,

Magnitude of acceleration of the electron, *${a}_{\mathrm{e}}=\frac{{q}_{\mathrm{e}}E}{{m}_{\mathrm{e}}}$*

Magnitude of acceleration of the proton, *${a}_{\mathrm{p}}=\frac{{q}_{\mathrm{p}}E}{{m}_{\mathrm{p}}}$*

Let *t* be the time taken by the electron and proton to reach the positive and negative plates, respectively.

The initial velocities of the proton and electron are zero.

Thus, the distance travelled by the proton is given by

$x=\frac{1}{2}\frac{{q}_{\mathrm{p}}E}{{m}_{\mathrm{p}}}{t}^{2}$ ...(1)

And, the distance travelled by the electron is given by

$2-x=\frac{1}{2}\frac{{q}_{\mathrm{e}}E}{{m}_{\mathrm{e}}}{t}^{2}$ ...(2)

On dividing (1) by (2), we get

$\frac{x}{2-x}=\frac{\left({\displaystyle \frac{{q}_{\mathrm{p}}E}{{m}_{\mathrm{p}}}}\right)}{\left({\displaystyle \frac{{q}_{\mathrm{e}}E}{{m}_{\mathrm{e}}}}\right)}=\frac{{m}_{\mathrm{e}}}{{m}_{\mathrm{p}}}=\frac{9.1\times {10}^{-31}}{1.67\times {10}^{-27}}=5.449\times {10}^{-4}\phantom{\rule{0ex}{0ex}}\Rightarrow x=10.898\times {10}^{-4}-5.449\times {10}^{-4}x\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{10.898\times {10}^{-4}}{1.0005449}=1.08\times {10}^{-8}\mathrm{cm}$

#### Page No 167:

#### Question 24:

Convince yourself that parts (a), (b) and (c) figure (31-E13) are identical. Find the capacitance between the points *A* and* B* of the assembly.

FigureA B

#### Answer:

Parts (a), (b) and (c) are identical, as all of them form a bridge circuit. In that circuit, capacitors of 1 µF and 2 µF are connected to terminal A and the 5 µF capacitor and capacitors of 3 µF and 6 µF are connected to terminal B and the 5 µF capacitor.

For the given situation, it can be observed that the bridge is in balance; thus, no current will flow through the 5 µF capacitor.

So to simplify the circuit, 5 µF capacitor can be removed from the circuit.

Now, 1 µF and 3 µF capacitors are in series.

And 2 µF and 6 µF capacitors are also in series combination.

These two combination are in parallel with each other.

The equivalent capacitance can be calculated as:

${C}_{\mathrm{eq}}=\frac{1\times 3}{1+3}+\frac{2\times 6}{2+6}\phantom{\rule{0ex}{0ex}}=\frac{3}{4}+\frac{12}{8}=\frac{9}{4}\mathrm{\mu F}=2.25\mu F$

∴ *C*_{eq} = 2.25 µF

Thus, parts (a), (b) and (c) are identical.

And,

*C*_{eq} = 2.25 µF

#### Page No 167:

#### Question 25:

Find the potential difference ${V}_{a}-{V}_{b}$ between the points *a* and *b* shown in each part of the figure (31-E14).

Figure

#### Answer:

(a) $q={q}_{1}+{q}_{2}$ ...(i)

On applying Kirchhoff's voltage law in the loop CabDC, we get

$\frac{{q}_{2}}{2}+\frac{{q}_{2}}{4}-\frac{{q}_{1}}{4}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{q}_{2}+{q}_{2}-{q}_{1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{q}_{2}={q}_{1}...\left(\mathrm{ii}\right)$

On applying Kirchhoff's voltage law in the loop DCBAD, we get

$\frac{q}{2}+\frac{{q}_{1}}{4}-12=0$

$\Rightarrow \frac{{q}_{1}+{q}_{2}}{2}+\frac{{q}_{1}}{4}-12=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{q}_{1}+2{q}_{2}=48...\left(\mathrm{iii}\right)$

From eqs. (ii) and (iii), we get

$9{q}_{2}+2{q}_{2}=48\phantom{\rule{0ex}{0ex}}\Rightarrow 11{q}_{2}=48\phantom{\rule{0ex}{0ex}}\Rightarrow {q}_{2}=\frac{48}{11}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{a}-{\mathrm{V}}_{b}=\frac{{q}_{2}}{4\mathrm{\mu F}}=\frac{48}{44}=\frac{12}{11}\mathrm{V}$

(b) Let the charge in the loop be *q*.

Now, on applying Kirchhoff's voltage law in the loop, we get

$\frac{q}{2}+\frac{q}{4}-24+12=0$

$\Rightarrow \frac{3q}{4}=12\phantom{\rule{0ex}{0ex}}\Rightarrow q=16\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{V}_{a}-{V}_{b}=\frac{-q}{2\mathrm{\mu F}}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{a}-{V}_{b}=\frac{-16\mathrm{\mu C}}{2\mathrm{\mu F}}=-8\mathrm{V}$

(c) ${V}_{\mathrm{a}}-{V}_{\mathrm{b}}=2-\frac{2-q}{2\mathrm{\mu F}}$

In the loop,

$2+2-\frac{q}{2}-\frac{q}{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow q=4\mathrm{C}$

$\therefore {V}_{\mathrm{a}}-{V}_{\mathrm{b}}=2-\frac{4}{2}=2-2=0\mathrm{V}$

(d)

Net charge flowing through all branches, *q* = 24 + 24 + 24 = 72 μC

Net capacitance of all branches, *C* = 4 + 2 + 1 = 7 μF

The total potential difference (*V*) between points *a* and *b* is given by

$V=\frac{q}{C}\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{72}{7}=10.3\mathrm{V}$

As the negative terminals of the batteries are connected to *a*, the net potential between points *a* and *b *is $-$10.3 V.

#### Page No 167:

#### Question 26:

Find the equivalent capacitances of the combinations shown in figure (31-E15) between the indicated points.

Figure

#### Answer:

(a)

Applying star-delta conversion in the part indicated in the diagram.

The capacitance of the C_{1} is given by

${C}_{1}\text{'}=\frac{{C}_{2}{C}_{3}}{{C}_{1}+{C}_{2}+{C}_{3}}\phantom{\rule{0ex}{0ex}}{C}_{2}\text{'}=\frac{{C}_{1}{C}_{3}}{{C}_{1}+{C}_{2}+{C}_{3}}\phantom{\rule{0ex}{0ex}}{C}_{3}\text{'}=\frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}+{C}_{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{1}\text{'}=\frac{3\times 4}{1+3+4}=\frac{12}{8}\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{2}\text{'}=\frac{1\times 3}{1+3+4}=\frac{3}{8}\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{3}\text{'}=\frac{1\times 4}{1+3+4}=\frac{4}{8}\mathrm{\mu F}$

Thus, the equivalent circuit can be drawn as:

Therefore, the equivalent capacitance is given by

C_{eq} = $\frac{3}{8}+\left[\frac{\left(3+{\displaystyle \frac{1}{2}}\right)\times \left({\displaystyle \frac{3}{2}}+1\right)}{\left(3+{\displaystyle \frac{1}{2}}\right)+\left({\displaystyle \frac{3}{2}}+1\right)}\right]=\frac{3}{8}+\frac{35}{24}=\frac{9+35}{24}=\frac{11}{6}\mathrm{\mu F}$

(b) By star-delta conversion,

$1+1=2\mu \mathrm{F}$

(c)

It is a balanced bridge.

Therefore, the capacitor of capacitance 5 μF can be removed.

The capacitance of the two branches are:

${C}_{1}=\frac{2\times 4}{2+4}=\frac{4}{3}\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}{C}_{2}=\frac{4\times 8}{4+8}=\frac{8}{3}\mathrm{\mu F}$

∴ Equivalent capacitance $=\frac{4}{3}+\frac{8}{3}+4$$=8\mu \mathrm{F}$

(d)

It is also a balanced diagram.

It can be observed that the bridges are balanced.

Therefore, the capacitors of capacitance 6 μF between the branches can be removed

The capacitances of the four branches are:

${C}_{1}=\frac{2\times 4}{2+4}=\frac{4}{3}\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}{C}_{2}=\frac{4\times 8}{4+8}=\frac{8}{3}\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{3}=\frac{4\times 8}{4+8}=\frac{8}{3}\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{4}=\frac{2\times 4}{2+4}=\frac{4}{3}\mathrm{\mu F}$

∴ Equivalent capacitance

$=\frac{4}{3}+\frac{8}{3}+\frac{8}{3}+\frac{4}{3}\phantom{\rule{0ex}{0ex}}=8\mathrm{\mu F}$

#### Page No 167:

#### Question 27:

Find the capacitance of the combination shown in figure (31-E16) between *A* and *B*.

Figure

#### Answer:

Capacitors 5 and 1 are in series.

Their equivalent capacitance, ${C}_{\mathrm{eq}}=\frac{{C}_{1}{C}_{5}}{{C}_{1}+{C}_{5}}=\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

*∴ **C*_{eq} = 1

Now, this capacitor system is parallel to capacitor 6. Thus, the equivalent capacitance becomes

1 + 1 = 2 μF

The above capacitor system is in series with capacitor 2. Thus, the equivalent capacitance becomes

$\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

The above capacitor system is in parallel with capacitor 7. Thus, the equivalent capacitance becomes

1 + 1 = 2 μF

The above capacitor system is in series with capacitor 3. Thus, the equivalent capacitance becomes

$\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

The above capacitor system is in parallel with capacitor 8. Thus, the equivalent capacitance becomes

1 + 1 = 2 μF

The above capacitor system is in series with capacitor 4. Thus, the equivalent capacitance becomes

$\frac{2\times 2}{2+2}=1\mathrm{\mu F}$

Hence, the equivalent capacitance between points A and B of the given capacitor system is 1 μF.

#### Page No 167:

#### Question 28:

Find the equivalent capacitance of the infinite ladder shown in figure (31-E17) between the points *A* and *B*.

Figure

#### Answer:

Let the equivalent capacitance of the infinite ladder be *C*.

Because it is an infinite ladder, the change in the equivalent capacitance will be negligible if we add one more ladder at point AB, as shown in the given figure.

From the given figure, the equivalent capacitance can be calculated as:

${C}_{eq}=\frac{2\times C}{2+C}+1=C\phantom{\rule{0ex}{0ex}}\Rightarrow \left(2+C\right)C=3C+2\phantom{\rule{0ex}{0ex}}\Rightarrow 2C+{C}^{2}=3C+2\phantom{\rule{0ex}{0ex}}\Rightarrow {C}^{2}-C-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(C-2\right)\left(C+1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow C=-1\mathrm{or}C=2\phantom{\rule{0ex}{0ex}}\Rightarrow C=-1\left(\mathrm{It}\mathrm{is}\mathrm{not}\mathrm{possible}.\right)\phantom{\rule{0ex}{0ex}}\therefore C=2\mathrm{\mu F}$

#### Page No 167:

#### Question 29:

A finite ladder is constructed by connecting several sections of 2 µF, 4 µF capacitor combinations as shown in figure (31-E18). It is terminated by a capacitor of capacitance *C*. What value should be chosen for *C*, such that the equivalent capacitance of the ladder between the points *A* and *B* becomes independent of the number of sections in between ?

Figure

#### Answer:

The equivalent capacitance of the ladder between points *A* and *B* becomes independent of the number of sections in between when the capacitance between A and B is *C*.

The capacitors *C* and 4 µF are in series; their equivalent capacitance is given by

${C}_{1}=\frac{C\times 4}{C+4}$

The capacitors *C*_{1} and 2 µF are in parallel; their equivalent capacitance is given by

*C = **C*_{1} + 2 µF

$\Rightarrow C=\frac{C\times 4}{C+4}+2\phantom{\rule{0ex}{0ex}}$

$\Rightarrow $4*C* + 8 + 2* C* = 4*C** *+ *C*^{2}

$\Rightarrow $*C*^{2} − 2*C* − 8 = 0

$\Rightarrow $*C *= −2, C = 4

Capacitance cannot be negative.

∴ *C* = 4 µF

The value of *C* is 4 µF.

#### Page No 167:

#### Question 30:

A charge of $+2\xb70\times {10}^{-8}\mathrm{C}$ is placed on the positive plate and a charge of $-1\xb70\times {10}^{-8}\mathrm{C}$ on the negative plate of a parallel-plate capacitor of capacitance $1\xb72\times {10}^{-3}\mu \mathrm{F}$. Calculate the potential difference developed between the plates.

#### Answer:

The charge on the positive plate is *q*_{1}_{ }and that on the negative plate is *q*_{2}.

Given:

${q}_{1}=2.0\times {10}^{-8}\mathrm{C}\phantom{\rule{0ex}{0ex}}{q}_{2}=-1.0\times {10}^{-8}\mathrm{C}\phantom{\rule{0ex}{0ex}}$

Now,

Net charge on the capacitor = $\frac{\left({q}_{1}-{q}_{2}\right)}{2}=1.5\times {10}^{-8}\mathrm{C}$

The potential difference developed between the plates is given by

$q=VC\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{1.5\times {10}^{-8}}{1.2\times {10}^{-9}}=12.5\mathrm{V}$

#### Page No 168:

#### Question 31:

A charge of 20 µC is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10 µF. Calculate the potential difference developed between the plates.

#### Answer:

Given:

Capacitance of the isolated capacitor = 10 µF

Charge on the positive plate = 20 µC

Effective charge on the capacitor = $\frac{20-0}{2}=10\mathrm{\mu C}$

The potential difference between the plates of the capacitor is given by

$V=\frac{Q}{C}$

∴ Potential difference = $\frac{10\mathrm{\mu C}}{10\mathrm{\mu F}}=1\mathrm{V}$

#### Page No 168:

#### Question 32:

A charge of 1 µC is given to one plate of a parallel-plate capacitor of capacitance 0⋅1 µF and a charge of 2 µC is given to the other plate. Find the potential difference developed between the plates.

#### Answer:

Charges on the plates:

${q}_{1}=1\mathrm{\mu C}=1\times {10}^{-6}\mathrm{C}\mathit{}\phantom{\rule{0ex}{0ex}}{q}_{2}=2\mathrm{\mu C}=2\times {10}^{-6}\mathit{}\mathrm{C}\phantom{\rule{0ex}{0ex}}$

The effective charge on the capacitor is given by

${q}_{\mathrm{net}}=\frac{{q}_{2}-{q}_{1}}{2}=\frac{(2-1)\times {10}^{-6}}{2}=0.5\times {10}^{-6}\mathrm{C}$

The potential difference is given by

$V=\frac{{q}_{\mathrm{net}}}{C}=\frac{0.5\times {10}^{-6}}{0.1\times {10}^{-6}}=5\mathrm{V}$

#### Page No 168:

#### Question 33:

Each of the plates shown in figure (31-E19) has surface area $\left(96/{\epsilon}_{0}\right)\times {10}^{-12}$ Fm on one side and the separation between the consecutive plates is 4⋅0 mm. The emf of the battery connected is 10 volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.

Figure

#### Answer:

Here, three capacitors are formed and each of them has area *A.*

Now,

$A=\frac{96}{{\in}_{0}}\times {10}^{-12}\mathrm{F}-\mathrm{m}\phantom{\rule{0ex}{0ex}}d=4\mathrm{mm}\phantom{\rule{0ex}{0ex}}=3\times {10}^{-3}\mathrm{m}$

The capacitance of the capacitors is given by

$C=\frac{{\in}_{0}A}{d}=\frac{{\in}_{0}\left(96/{\in}_{0}\right)\times {10}^{-12}}{4\times {10}^{-3}}=24\times {10}^{-9}\mathrm{F}$

Because the three capacitors are arranged in series, the equivalent capacitance is given by

${C}_{\mathrm{eq}}=\frac{24\times {10}^{-9}}{3}=8\times {10}^{-9}\mathrm{F}$

The total charge on the capacitor is given by

*Q* = *C*_{eq}*V*

$\Rightarrow Q=8\times {10}^{-9}\times 10\phantom{\rule{0ex}{0ex}}\Rightarrow Q=8\times {10}^{-8}C$

The charge on a single plate is given by

${Q}_{\mathrm{P}}=2\times 8\times {10}^{-8}\phantom{\rule{0ex}{0ex}}\Rightarrow {Q}_{\mathrm{P}}=16\times {10}^{-8}=0.16\mathrm{\mu C}$

#### Page No 168:

#### Question 34:

The capacitance between the adjacent plates shown in figure (31-E20) is 50 nF. A charge of 1⋅0 µC is placed on the middle plate. (a) What will be the charge on the outer surface of the upper plate? (b) Find the potential difference developed between the upper and the middle plates.

Figure

#### Answer:

(a) When the charge of 1 µC is introduced on plate B, the charge gets equally distributed on its surface. All sides of the plate gets 0.5 µC of charge. Due to induction, 0.5 µC charge is induced on the upper surface of plate A.

(b) Given:

Capacitance, *C* = 50 nF = 50 × 10^{$-$9} F = 5 × 10^{$-$8} F

Effective charge on the capacitor, ${Q}_{\mathrm{net}}=\frac{1\mathrm{\mu C}-0}{2}=0.5\mathrm{\mu C}$

∴ Potential difference between plates B and A, $V=\frac{{Q}_{\mathrm{net}}}{C}=\frac{5\times {10}^{-7}}{5\times {10}^{-8}}=10\mathrm{V}$

#### Page No 168:

#### Question 35:

Consider the situation of the previous problem. If 1⋅0 µC is placed on the upper plate instead of the middle, what will be the potential difference between (a) the upper and the middle plates and (b) the middle and the lower plates?

#### Answer:

When 1µC charge is given to the upper plate, the charge gets distributed. The two sides of the upper plate have +0.5 µC charge. One side of the middle plate facing the upper plate has $-$0.5 µC charge and the other side has +0.5 µC charge. Similarly, one side of the lower plate facing the middle plate has $-$0.5 µC charge and the other side has +0.5 µC charge.

(a) Effective charge on the capacitor formed by the upper and middle plates = 0.5 µC

Capacitance = 50 nF = 0.05 µF

∴ Potential difference between the plates, $V=\frac{Q}{C}=\frac{0.5\mathrm{\mu C}}{0.05\mathrm{\mu F}}=10\mathrm{V}$

(b) Effective charge on the capacitor formed by the middle and lower plates = 0.5 µC

Capacitance = 0.50 µF

∴ Potential difference between the plates, $V=\frac{Q}{C}=\frac{0.5\mathrm{\mu C}}{0.05\mathrm{\mu F}}=10\mathrm{V}$

#### Page No 168:

#### Question 36:

Two capacitors of capacitance 20⋅0 pF and 50⋅0 pF are connected in series with a 6⋅00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

#### Answer:

Given:

${\mathrm{C}}_{1}=20.0\mathrm{pF}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{2}=50.0\mathrm{pF}$

When the capacitors are connected in series, their equivalent capacitance is given by

${\mathrm{C}}_{\mathrm{eq}}=\frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}$

∴ Equivalent capacitance, ${\mathrm{C}}_{\mathrm{eq}}=\frac{(50\times {10}^{-12})\times (20\times {10}^{-12})}{(50\times {10}^{-12})+(20\times {10}^{-12})}=1.428\times {10}^{-11}\mathrm{F}$

(a) The charge on both capacitors is equal as they are connected in series. It is given by

$q={C}_{\mathrm{eq}}\times V\phantom{\rule{0ex}{0ex}}\Rightarrow q=(1.428\times {10}^{-11})\times 6.0\mathrm{C}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{V}_{1}=\frac{q}{{\mathrm{C}}_{1}}=\frac{(1.428\times {10}^{-11})\times 6.0\mathrm{C}}{(20\times {10}^{-12})}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{1}=4.29\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}{V}_{2}=\left(6.00-4.29\right)\mathrm{V}=1.71\mathrm{V}\phantom{\rule{0ex}{0ex}}$

(b) The energies in the capacitors are given by

${E}_{1}=\frac{{q}^{2}}{2{\mathrm{C}}_{1}}\phantom{\rule{0ex}{0ex}}={\left[(1.428\times {10}^{-11})\times 6.0\right]}^{2}\times \frac{1}{2\times 20\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}=184\mathrm{pJ}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}{E}_{2}=\frac{{q}^{2}}{2{\mathrm{C}}_{1}}\phantom{\rule{0ex}{0ex}}={\left[(1.428\times {10}^{-11})\times 6.0\right]}^{2}\times \frac{1}{2\times 50\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}=73.5\mathrm{pJ}$

#### Page No 168:

#### Question 37:

Two capacitors of capacitances 4⋅0 µF and 6⋅0 µF are connected in series with a battery of 20 V. Find the energy supplied by the battery.

#### Answer:

Given:

${C}_{1}=4\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}{C}_{2}=6\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}V=20\mathrm{V}$

Now,

The equivalent capacitance is given by

${C}_{\mathrm{eq}}=\frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}=\frac{4\times 6}{4+6}=2.4\mathrm{\mu F}$

The energy supplied by the battery is given by

$E={C}_{\mathrm{eq}}{V}^{2}\phantom{\rule{0ex}{0ex}}=\left(2.4\mathrm{\mu F}\right)\times {\left(20\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(2.4\mathrm{\mu F}\right)\times 400=960\mathrm{\mu J}$

#### Page No 168:

#### Question 38:

Each capacitor in figure (31-E21) has a capacitance of 10 µF. The emf of the battery is 100 V. Find the energy stored in each of the four capacitors.

Figure

#### Answer:

Capacitors *b* and *c* are in parallel; their equivalent capacitance is 20 µF.

Thus, the net capacitance of the circuit is given by

$\frac{1}{{C}_{\mathrm{net}}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{{C}_{\mathrm{net}}}=\frac{2+1+2}{20}=\frac{5}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{net}}=4\mathrm{\mu F}$

The total charge of the battery is given by

$Q={C}_{\mathrm{net}}V=(4\mathrm{\mu F})\times (100\mathrm{V})=4\times {10}^{-4}\mathrm{C}$

For *a* and *d*,

$q=4\times {10}^{-4}\mathrm{C}\mathrm{and}\mathrm{C}={10}^{-5}\mathrm{F}\phantom{\rule{0ex}{0ex}}\therefore E=\frac{{q}^{2}}{2C}=\frac{4\times {10}^{-4}}{2\times {10}^{-5}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=8\times {10}^{-3}\mathrm{J}=8\mathrm{mJ}$

For *b* and *c*,

$q=4\times {10}^{-4}\mathrm{C}\mathrm{and}{C}_{eq}=2C=2\times {10}^{-5}\mathrm{F}\phantom{\rule{0ex}{0ex}}\therefore V=\frac{q}{{C}_{eq}}\frac{4\times {10}^{-4}}{2\times {10}^{-5}}=20\mathrm{V}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1}{2}C{V}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1}{2}\times {10}^{-5}\times 400\phantom{\rule{0ex}{0ex}}\Rightarrow E=2\times {10}^{-3}\mathrm{J}=2\mathrm{mJ}$

#### Page No 168:

#### Question 39:

A capacitor with stored energy 4⋅0 J is connected with an identical capacitor with no electric field in between. Find the total energy stored in the two capacitors.

#### Answer:

Given:

Energy stored in the charged capacitor = 4.0 J

When the capacitors are connected, the charge flows from the charged capacitor to the uncharged capacitor. Because the capacitors are identical, the charge flows till the charge in both the capacitors becomes equal.

The energy of a capacitor is given by

$E=\frac{{q}^{2}}{2C}$

As the charge in both the capacitors is the same, their capacitance is also the same. So, the energy is equally divided between them.

Thus, the energy stored in each of the capacitors is 2.0 J.

#### Page No 168:

#### Question 40:

A capacitor of capacitance 2⋅0 µF is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance 4⋅0 µF as shown in figure (31-E22). Find (a) the charge on each of the two capacitors after the connection, (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other.

Figure

#### Answer:

Charge on the 2 µF capacitor when it is not connected to the 4 µF capacitor is given by

$q\mathit{=}C\mathit{\times}V=12\times 2\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow q=24\times {10}^{-6}\mathrm{C}$

(a) On connecting the capacitors, the charge flows from the 2 µF capacitor to the 4 µF capacitor.

Now, let the charges on the 2 µF and 4 µF capacitors be *q*_{1} and *q*_{2}, respectively.

As they are connected in parallel, the potential difference across them is the same.

$\therefore V=\frac{{q}_{1}}{{C}_{1}}=\frac{{q}_{2}}{{C}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{{q}_{1}}{2}=\frac{{q}_{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {q}_{2}=2{q}_{1}$

The total charge on the capacitors will be the same as the initial charge stored on the 2 µF capacitor.

$\therefore {q}_{1}+{q}_{2}=24\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow 3{q}_{1}=24\times {10}^{-6}\mathrm{C}\phantom{\rule{0ex}{0ex}}\Rightarrow {q}_{1}=8\times {10}^{-6}\mathrm{C}=8\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}\Rightarrow {q}_{1}=2{q}_{1}=16\mathrm{\mu C}$

(b) Energies stored in the capacitors are given by

${E}_{1}=\frac{1}{2}\times \frac{{{q}_{1}}^{2}}{{C}_{1}}=16\mathrm{\mu J}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}{E}_{2}=\frac{1}{2}\times \frac{{{q}_{2}}^{2}}{{C}_{2}}=32\mathrm{\mu J}$

(c)

Initial energy stored in the 2 µF capacitor is given by

${E}_{\mathrm{i}}=\frac{1}{2}C{V}^{2}=\frac{1}{2}\times \left(2\times {10}^{-6}\right){\left(12\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{\mathrm{i}}=144\mathrm{\mu J}$

Total energy of the capacitors after they are connected in parallel is given by

*E*_{f} = *E*_{1} + *E*_{2}

$\Rightarrow $*E*_{f} = 16 + 32 = 48 μJ

Heat produced during the charge transfer is given by

*E* = *E*_{f} $-$ *E*_{i}

$\Rightarrow $*E* = 144 $-$ 48 = 96 μJ

#### Page No 168:

#### Question 41:

A point charge *Q* is placed at the origin. Find the electrostatic energy stored outside the sphere of radius *R* centred at the origin.

#### Answer:

Given:

Charge on the sphere = *Q*

Radius of the sphere = *R*

Capacitance of the sphere, $C=4\pi {\in}_{0}R$

Thus, the energy of the sphere is given by

$E=\frac{1}{2}\mathrm{C}{V}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 4\pi {\in}_{0}R\times \frac{{Q}^{2}}{{\left(4\pi {\in}_{0}R\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{Q}^{2}}{8\pi {\in}_{0}R}$

#### Page No 168:

#### Question 42:

A metal sphere of radius *R* is charged to a potential *V*. (a) Find the electrostatic energy stored in the electric field within a concentric sphere of radius 2 *R*. (b) Show that the electrostatic field energy stored outside the sphere of radius 2 *R* equals that stored within it.

#### Answer:

Potential of the inner metallic sphere* *is given by

$V=\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{q}{R}$

Capacitance of the capacitor formed by two concentric spheres of radii *R* and 2*R* is given by

$C=4\pi {\in}_{0}\frac{R\times 2R}{2R-R}\phantom{\rule{0ex}{0ex}}\Rightarrow C=4\pi {\in}_{0}\times 2R$

Potential of the outer sphere is given by

$\mathit{}{V}_{2}=\frac{1}{4\mathrm{\pi}{\in}_{0}}\frac{q}{2R}=\frac{V}{2}$

Electrostatic energy stored outside the sphere is given by

$E=\frac{1}{2}C{\left(V-{V}_{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{1}{2}\times 4\mathrm{\pi}{\in}_{0}\times 2R\times \frac{{V}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\mathrm{\pi}{\in}_{0}\times R{V}^{2}$

#### Page No 168:

#### Question 43:

A large conducting plane has a surface charge density $1\xb70\times {10}^{-4}\mathrm{C}{\mathrm{m}}^{-2}$. Find the electrostatic energy stored in a cubical volume of edge 1⋅0 cm in front of the plane.

#### Answer:

Given,

Surface charge density of the plane, $\sigma =1.0\times {10}^{-4}\mathrm{C}/{\mathrm{m}}^{2}$

Volume of the cube, $V={a}^{3}={10}^{-6}{\mathrm{m}}^{3}$

Electric field near the charged conducting plane is given as,

$E=\frac{\sigma}{{\in}_{0}}...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}$

Energy density of electric filed,

$u=\frac{1}{2}{\in}_{0}{E}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{1}{2}{\in}_{0}{\left(\frac{\sigma}{{\in}_{0}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{1}{2}\frac{{\sigma}^{2}}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}\Rightarrow u=\frac{1}{2}\times \frac{(1.0\times {10}^{-4}{)}^{2}}{8.85\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}\Rightarrow u=0.056\times {10}^{4}j/{m}^{3}\phantom{\rule{0ex}{0ex}}volume={10}^{-6}{m}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow U=u\times V\phantom{\rule{0ex}{0ex}}\Rightarrow U=0.056\times {10}^{4}\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow U=5.6\times {10}^{-4}J$

#### Page No 168:

#### Question 44:

A parallel-plate capacitor having plate area 20 cm^{2} and separation between the plates 1⋅00 mm is connected to a battery of 12⋅0 V. The plates are pulled apart to increase the separation to 2⋅0 mm. (a) Calculate the charge flown through the circuit during the process. (b) How much energy is absorbed by the battery during the process? (c) Calculate the stored energy in the electric field before and after the process. (d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart. (e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.

#### Answer:

Given:

$\mathrm{Area},\mathit{}A=20{\mathrm{cm}}^{2}=2\times {10}^{-3}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Separation},{d}_{1}=1\mathrm{mm}={10}^{-3}\mathrm{m}$

Initial capacitance of the capacitor:

${C}_{1}=\frac{{\in}_{0}A}{{d}_{1}}\phantom{\rule{0ex}{0ex}}{C}_{1}=\frac{8.88\times {10}^{-12}\times 20\times {10}^{-4}}{1\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}{C}_{1}=1.776\times {10}^{-11}\mathrm{F}$

Final capacitance of the capacitor:

${C}_{2}=\frac{{C}_{1}}{2}\left(\mathrm{because}{d}_{2}=\frac{{d}_{1}}{2}\right)$

(a) Charge flown through the circuit:

$Q={C}_{1}V-{\mathrm{C}}_{2}V\phantom{\rule{0ex}{0ex}}\mathrm{Q}=\left({C}_{1}-{C}_{2}\right)V\phantom{\rule{0ex}{0ex}}Q=\frac{1.776}{2}\times {10}^{-11}\times 12.0\phantom{\rule{0ex}{0ex}}Q=1.06\times {10}^{-10}\mathrm{C}$

(b) Energy absorbed by the battery:

$E=QV=1.06\times {10}^{-10}\times 12\phantom{\rule{0ex}{0ex}}E=12.72\times {10}^{-10}\mathrm{J}$

(c) Energy stored before the process, *E*_{i}$=\frac{1}{2}{C}_{1}{V}^{2}$

$\Rightarrow {E}_{\mathrm{i}}=\frac{1}{2}\times 1.776\times {10}^{-11}\times {\left(12\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{\mathrm{i}}=12.7\times {10}^{-10}\mathrm{J}$

Energy stored after the process, *E*_{f}$=\frac{1}{2}{C}_{2}{V}^{2}$

$\Rightarrow {E}_{\mathrm{f}}=\frac{1}{2}\times \frac{1.776}{2}\times {10}^{-11}\times {\left(12\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{\mathrm{f}}=6.35\times {10}^{-10}\mathrm{J}$

(d) Work done, *W* = Force × Distance

*W* = Change in the energy

*$\Rightarrow $W* = 6.35 × 10^{−10} J

#### Page No 168:

#### Question 45:

A capacitor having a capacitance of 100 µF is charged to a potential difference of 24 V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12 V with the positive plate of the capacitor joined with the positive terminal of the battery. (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the 12 V battery. (c) Is work done by the battery or is it done on the battery? Find its magnitude. (d) Find the decrease in electrostatic field energy. (e) Find the heat developed during the flow of charge after reconnection.

#### Answer:

(a) Before connecting to the battery of 12 volts,

$C=100\mu \mathrm{F}\mathrm{and}V=24\mathrm{V}\phantom{\rule{0ex}{0ex}}\therefore q=CV=2400\mathrm{\mu F}$

After connecting to the battery of 12 volts,

$C=100\mu \mathrm{F}\mathrm{and}V=12\mathrm{V}\phantom{\rule{0ex}{0ex}}\therefore q=CV=1200\mathrm{\mu C}$

(b) Charge flown through the 12 V battery is 1200 $\mathrm{\mu C}$.

(c) We know,

$W=Vq=12\times 1200\phantom{\rule{0ex}{0ex}}=14400\mathrm{J}\phantom{\rule{0ex}{0ex}}=14.4\mathrm{mJ}$

It is the work done on the battery.

(d) Initial electrostatic field energy:

${U}_{\mathrm{i}}=\frac{1}{2}C{V}_{1}^{2}$

Final electrostatic field energy:

${U}_{\mathrm{f}}=\frac{1}{2}C{V}_{2}^{2}$

Decrease in the electrostatic field energy:

$\u2206U=\frac{1}{2}\left(\mathrm{C}{V}_{1}^{2}-{\mathrm{CV}}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\mathrm{C}\left({V}_{1}^{2}-{V}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}=100\left[{\left(24\right)}^{2}-{\left(12\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 100\times \left(576-144\right)\phantom{\rule{0ex}{0ex}}=21600\mathrm{J}=21.6\mathrm{mJ}$

(e) After reconnection,

$\mathrm{C}=100\mathrm{\mu F}\mathrm{and}V=12\mathrm{V}$

Heat developed during the flow of charge after reconnection is given by

$H=\frac{1}{2}\mathrm{C}{V}^{2}=100\times 144\phantom{\rule{0ex}{0ex}}=7200\mathrm{J}=7.2\mathrm{mJ}$

This amount of energy is developed as heat when the charge flows through the capacitor.

#### Page No 168:

#### Question 46:

Consider the situation shown in figure (31-E23). The switch *S* is open for a long time and then closed. (a) Find the charge flown through the battery when the switch *S* is closed. (b) Find the work done by the battery.

Figure

(c) Find the change in energy stored in the capacitors.

(d) Find the heat developed in the system.

#### Answer:

Since the switch is opened for a long time, the capacitor is in series.

∴ ${C}_{eq}=\frac{C}{2}$

When the switch is closed, the charge flown from the battery is given by

$q=\frac{\mathrm{C}}{2}\times \epsilon =\frac{\mathrm{C}\epsilon}{2}$

(b) Work done, *W*$=q\times V$

$\Rightarrow q=\frac{\mathrm{C}\epsilon}{2}\times \epsilon =\frac{\mathrm{C}{\epsilon}^{2}}{2}$

(c) The change in the energy stored in the capacitors is given by

$\u2206E=\frac{1}{2}{\mathrm{C}}_{eq}\times {v}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{\mathrm{C}}{2}\times {\epsilon}^{2}=\frac{1}{4}\mathrm{C}{\epsilon}^{2}$

(d) The heat developed in the system is given by

$H=\u2206E\phantom{\rule{0ex}{0ex}}\Rightarrow H=\frac{1}{4}\mathrm{C}{\epsilon}^{2}$

#### Page No 169:

#### Question 47:

A capacitor of capacitance 5⋅00 µF is charged to 24⋅0 V and another capacitor of capacitance 6⋅0 µF is charged to 12⋅0 V. (a) Find the energy stored in each capacitor. (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the new charges on the capacitors. (c) Find the loss of electrostatic energy during the process. (d) Where does this energy go?

#### Answer:

Given:

${C}_{1}=5\mathrm{\mu F}\mathrm{and}{V}_{1}=24\mathrm{V}\phantom{\rule{0ex}{0ex}}\therefore {q}_{1}={C}_{1}{V}_{1}=5\times 24=120\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}{C}_{2}=6\mathrm{\mu F}\mathrm{and}{V}_{2}=12\mathrm{V}\phantom{\rule{0ex}{0ex}}\therefore {q}_{2}={C}_{2}{V}_{2}=6\times 12=72\mathrm{\mu C}$

(a)

Energy stored in the first capacitor:

${U}_{1}=\frac{1}{2}{C}_{1}{V}_{1}^{2}\phantom{\rule{0ex}{0ex}}=1440\mathrm{J}=1.44\mathrm{mJ}$

Energy stored in the second capacitor:

${U}_{\mathit{2}}=\frac{1}{2}{\mathrm{C}}_{1}{\mathrm{V}}_{2}^{2}\phantom{\rule{0ex}{0ex}}=432\mathrm{J}=0.432\mathrm{mJ}$

(b) The capacitors are connected to each other in such a way that the positive plate of the first capacitor is connected to the negative plate of the second capacitor and vice versa.

∴ Net change in the system, *Q*_{net} = 120 $-$ 72 = 48

Now, let *V* be the common potential of the two capacitors.

From the conservation of charge before and after connecting, we get

$V=\frac{{Q}_{\mathrm{net}}}{({C}_{1}+{C}_{2})}\phantom{\rule{0ex}{0ex}}=\frac{48}{(5+6)}\phantom{\rule{0ex}{0ex}}=4.36\mathrm{V}$

New charges:

${q}_{1}\mathit{\text{'}}\mathit{=}{C}_{1}V=5\times 4.36=21.8\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}{q}_{2}\text{'}={C}_{2}V=6\times 4.36=26.2\mathrm{\mu C}$

(c)

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}{U}_{1}=\frac{1}{2}{C}_{1}{V}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}{U}_{2}=\frac{1}{2}{C}_{2}{V}^{2}\phantom{\rule{0ex}{0ex}}\therefore {U}_{f}=\frac{1}{2}{\mathrm{V}}^{2}\left({\mathrm{C}}_{1}+{\mathrm{C}}_{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\left(4.36\right)}^{2}\left(5+6\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 19\times 11\phantom{\rule{0ex}{0ex}}=104.5\times {10}^{-6}\mathrm{J}\phantom{\rule{0ex}{0ex}}=0.1045\mathrm{mJ}$

$\mathrm{But}{U}_{i}=1.44+0.433=1.873\phantom{\rule{0ex}{0ex}}\mathrm{Loss}\mathrm{of}\mathrm{energy}:\phantom{\rule{0ex}{0ex}}\u2206U=1.873-0.1045\phantom{\rule{0ex}{0ex}}=1.7678\phantom{\rule{0ex}{0ex}}=1.77\mathrm{mJ}$

(d) The energy is dissipated as heat.

#### Page No 169:

#### Question 48:

A 5⋅0 µF capacitor is charged to 12 V. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires.

#### Answer:

As a capacitor of capacitance *C* is connected to the battery of voltage *V*, one of its plates gets charged to *Q = CV* and the other gets charged to −*Q* coulomb. When the polarity is reversed, charge −*Q* appears on the first plate and charge +*Q* appears on the second plate. So to maintain these charges, charge 2*Q* passes through the battery from the negative terminal to the positive terminal.

The work done by the battery is given by

$W=2QV=2C{E}^{2}$

In the given process, the capacitor's energy remains the same in both cases. The work done by the battery appears as heat in the connecting wires.

Now,

Heat produced:

$H=2C{E}^{2}\phantom{\rule{0ex}{0ex}}H=2\times \left(5\times {10}^{-6}\right)\times 144\phantom{\rule{0ex}{0ex}}H=144\times {10}^{-5}\mathrm{J}=1.44\mathrm{mJ}$

#### Page No 169:

#### Question 49:

The two square faces of a rectangular dielectric slab (dielectric constant 4⋅0) of dimensions 20 cm × 20 cm × 1⋅0 mm are metal-coated. Find the capacitance between the coated surfaces.

#### Answer:

The area of the plates of the capacitor is given by

$A=20\mathrm{cm}\times 20\mathrm{cm}=400{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A=4\times {10}^{-2}\mathrm{m}$

The separation between the parallel plates is given by

$d=1\mathrm{m}=1\times {10}^{-3}\mathrm{m}$

Here, the thickness of the dielectric is the same as the separation between the parallel plates.

Thus, the capacitance is given by

$C=\frac{{\in}_{0}Ak}{d}=\frac{(8.85\times {10}^{-12})\times (4\times {10}^{-2})\times 4}{{10}^{-3}}=1.42\mathrm{nF}$

#### Page No 169:

#### Question 50:

If the above capacitor is connected across a 6⋅0 V battery, find (a) the charge supplied by the battery, (b) the induced charge on the dielectric and (c) the net charge appearing on one of the coated surfaces.

#### Answer:

Given:

Capacitance of the capacitor, *C* = 1.42 nF

Dielectric constant, *k* = 4

Voltage of the battery connected, *V* = 6 V

(a) The charge supplied by the battery is given by

$q=\mathrm{C}V=1.42\times {10}^{-9}\times 6\phantom{\rule{0ex}{0ex}}=8.52\times {10}^{-9}\mathrm{C}=8.5\mathrm{nC}$

(b) The charge induced on the dielectric is given by

$q\text{'}=q\left(1-\frac{1}{\mathrm{K}}\right)\phantom{\rule{0ex}{0ex}}=8.52\times {10}^{-9}\times \left(1-0.25\right)\phantom{\rule{0ex}{0ex}}=6.39\times {10}^{-9}=6.4\mathrm{nC}$

(c) The charge appearing on one of the coated plates is given by

*Q*_{net}_{ }= 8.5 $-$ 6.4 = 2.1 nC

#### Page No 169:

#### Question 51:

The separation between the plates of a parallel-plate capacitor is 0⋅500 cm and its plate area is 100 cm^{2}. A 0⋅400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.

#### Answer:

Given:

Area of the plate = 100 cm^{2}^{ }

Separation between the plates = 0.500 cm = 5 × 10^{$-$3} m

Thickness of the metal, *t* = 4 × 10^{$-$3} m

$\therefore C=\frac{{\in}_{0}A}{d-t+{\displaystyle \frac{t}{k}}}\phantom{\rule{0ex}{0ex}}$

Here,

*k *= Dielectric constant of the metal

*d* = Separation between the plates

*t* = Thickness of the metal

For the metal*, k* = $\infty $.

$\therefore C=\frac{{\in}_{0}A}{d-t}=\frac{(8.85\times {10}^{-12})\times {10}^{-12}}{(5-4)\times {10}^{-3}}=88\mathrm{pF}\phantom{\rule{0ex}{0ex}}$

Here, the capacitance is independent of the position of the metal. At any position, the net separation is (*d* − *t*).

#### Page No 169:

#### Question 52:

A capacitor stores 50 µC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100 µC flows through the battery. Find the dielectric constant of the material inserted.

#### Answer:

Given:

Initial charge on the capacitor = 50 µC

Now, let the dielectric constant of the material inserted be *k*.

As 100 µC of extra charge flows through the battery, the net charge on the capacitor becomes

50 + 100 = 150 µC

Now,

${C}_{1}=\frac{{q}_{1}}{V}=\frac{{\in}_{0}A}{d}...\left(i\right)\phantom{\rule{0ex}{0ex}}{C}_{2}=\frac{{q}_{2}}{V}=\frac{{\in}_{0}Ak}{d}...\left(ii\right)$

On dividing (*i*) by (*ii*), we get

$\frac{{C}_{1}}{{C}_{2}}=\frac{{q}_{1}}{{q}_{2}}=\frac{1}{k}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{50}{150}=\frac{1}{k}\phantom{\rule{0ex}{0ex}}\Rightarrow k=3$

Thus, the dielectric constant of the given material is 3.

#### Page No 169:

#### Question 53:

A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?

#### Answer:

For the given capacitor,

$C=5\mathrm{\mu F}\phantom{\rule{0ex}{0ex}}V=6\mathrm{V}\phantom{\rule{0ex}{0ex}}d=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m}$

(a) The charge on the positive plate is calculated using *q* = *CV.*

Thus,

$q=5\mathrm{\mu F}\times 6\mathrm{V}=30\mathrm{\mu C}$

(b) The electric field between the plates of the capacitor is given by

$E=\frac{V}{d}=3\times {10}^{3}\mathrm{V}/\mathrm{m}$

(c) Separation between the plates of the capacitor, *d* = 2×10^{$-$3} m

Dielectric constant of the dielectric inserted, *k* = 5

Thickness of the dielectric inserted, *t* = 1×10^{$-$3} m

Now,

Area of the plates of the capacitor

$\mathrm{C}=\frac{{\in}_{0}\mathrm{A}}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow 5\times {10}^{-6}=\frac{8.85\times {10}^{-12}\times A}{2\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {10}^{4}=8.85\times A\phantom{\rule{0ex}{0ex}}\Rightarrow A={10}^{4}\times \frac{1}{8.85}$

When the dielectric placed in it, the capacitance becomes

${C}_{1}=\frac{{\in}_{0}A}{d-t+{\displaystyle \frac{t}{k}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{1}=\frac{8.85\times {10}^{-12}\times {\displaystyle \frac{{10}^{4}}{8.85}}}{2\times {10}^{-3}-{10}^{-3}+{\displaystyle \frac{{10}^{-3}}{5}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{1}=\frac{8.85\times {10}^{-12}\times {\displaystyle \frac{{10}^{4}}{8.85}}}{{10}^{-3}+{\displaystyle \frac{{10}^{-3}}{5}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{1}=\frac{{10}^{-12}\times {10}^{4}\times 5}{6\times {10}^{-3}}=8.33\mathrm{\mu F}$

(d)

The charge stored in the capacitor initially is given by

$C=5\times {10}^{-6}\mathrm{F}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\phantom{\rule{0ex}{0ex}}V=6\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}Q=CV=3\times {10}^{-5}\mathrm{F}\phantom{\rule{0ex}{0ex}}\Rightarrow Q=30\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}$

The charge on the capacitor after inserting the dielectric slab is given by

${C}_{1}=8.3\times {10}^{-6}\mathrm{F}\phantom{\rule{0ex}{0ex}}\therefore Q\text{'}={C}_{1}V=8.3\times 6\times {10}^{-6}\phantom{\rule{0ex}{0ex}}\Rightarrow Q\text{'}=50\mathrm{\mu C}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Charge}\mathrm{flown},Q\text{'}-Q=20\mathrm{\mu C}$

#### Page No 169:

#### Question 54:

A parallel-plate capacitor has plate area 100 cm^{2} and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.

#### Answer:

The given system of the capacitor will behave as two capacitors connected in series.

Let the capacitances be *C*_{1} and *C*_{2}.

Now,

${C}_{1}=\frac{{\in}_{0}A{k}_{1}}{{d}_{1}}\mathrm{and}{C}_{2}=\frac{{\in}_{0}A{k}_{2}}{{d}_{2}}$

Thus, the net capacitance is given by

$C=\frac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{{\in}_{0}A{k}_{1}}{{d}_{1}}\times \frac{{\in}_{0}A{k}_{2}}{{d}_{2}}}{\frac{{\in}_{0}A{k}_{1}}{{d}_{1}}+\frac{{\in}_{0}A{k}_{2}}{{d}_{2}}}\phantom{\rule{0ex}{0ex}}=\frac{{\in}_{0}A\left({k}_{1}+{k}_{2}\right)}{{k}_{1}{d}_{2}+{k}_{2}{d}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{(8.85\times {10}^{-12})\times \left({10}^{-2}\right)\times 24}{(6\times 4\times {10}^{-3}+4\times 6\times {10}^{-3})}=4.425\times {10}^{-11}\mathrm{C}\phantom{\rule{0ex}{0ex}}=44.25\mathrm{pF}$

#### Page No 169:

#### Question 55:

A parallel-plate capacitor having plate area 400 cm^{2} and separation between the plates 1⋅0 mm is connected to a power supply of 100 V. A dielectric slab of thickness 0⋅5 mm and dielectric constant 5⋅0 is inserted into the gap. (a) Find the increase in electrostatic energy. (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out?

#### Answer:

The capacitance of the capacitor without the dielectric slab is given by

${C}_{1}=\frac{{\in}_{0}A}{d}$

$\Rightarrow C=\frac{(8.85\times {10}^{-12})\times (400\times {10}^{-4})}{(1.0\times {10}^{-3})}\phantom{\rule{0ex}{0ex}}\Rightarrow C=3.54\times {10}^{-10}\mathrm{F}$

When the dielectric slab is inserted, the capacitance becomes

$C\text{'}=\frac{{\in}_{0}\mathit{}A}{\left(d-{\displaystyle t+\frac{t}{K}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow C\text{'}=\frac{(8.85\times {10}^{-12})\times (400\times {10}^{-4})}{\left(1\times {10}^{-3}-0.5\times {10}^{-3}+{\displaystyle \frac{0.5\times {10}^{-3}}{5}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow C\text{'}=\frac{5\times (8.85\times {10}^{-12})\times (400\times {10}^{-4})}{(6\times 0.5\times {10}^{-3})}\phantom{\rule{0ex}{0ex}}\Rightarrow C\text{'}=5.9\times {10}^{-10}\mathrm{F}$

(a) Increased electrostatic energy:

$\u2206E=\frac{1}{2}C\text{'}{V}^{2}-\frac{1}{2}\mathrm{C}{V}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206E=\frac{1}{2}\left(5.9-3.54\right)\times {10}^{-10}\times {\left(100\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206E=1.18\times {10}^{-6}\mathrm{J}=1.18\mathrm{\mu J}$

(b)

Charge on the capacitor containing dielectric is,

$Q\text{'}=C\text{'}V\phantom{\rule{0ex}{0ex}}\Rightarrow Q\text{'}=5.9\times {10}^{-10}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow Q\text{'}=5.9\times {10}^{-8}\mathrm{C}$

Potential difference across the capacitor after removing battery and then dielectric is,

$V\text{'}=\frac{Q\text{'}}{C}\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=\frac{5.9\times {10}^{-8}}{3.54\times {10}^{-10}}\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=166\mathrm{V}$

Increased electrostatic energy,

$\u2206E\text{'}=\frac{1}{2}CV{\text{'}}^{2}-\frac{1}{2}\mathrm{C}\text{'}{V}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206E\text{'}=\frac{1}{2}\left[3.54\times {10}^{-10}\times (166{)}^{2}-5.9\times {10}^{-10}\times (100{)}^{2}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206E\text{'}=1.92\times {10}^{-6}\mathrm{J}=1.92\mathrm{\mu J}$

(c) When the battery is connected, energy is increased after insertion of the dielectric slab because of the increase in the capacitance of the capacitor. Now, capacitor will abstract more change from the battery.

When the battery is disconnected, and dielectric slab is taken out then energy stored in the will increase because of increase of potential difference across the capacitor.

#### Page No 169:

#### Question 56:

Find the capacitances of the capacitors shown in figure (31-E24). The plate area is *A* and the separation between the plates is *d*. Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is filled with the dielectric slabs.

Figure

#### Answer:

The two parts of the capacitor are in series with capacitances *C*_{1}_{ }and *C*_{2}.

Here,

${C}_{1}=\frac{{K}_{1}{\in}_{0}A}{{\displaystyle \frac{d}{2}}}\mathrm{and}{C}_{2}=\frac{{K}_{2}{\in}_{0}A}{{\displaystyle \frac{d}{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{1}=\frac{2{K}_{1}{\in}_{0}A}{d}\mathrm{and}{C}_{2}=\frac{2{K}_{2}{\in}_{0}A}{d}$

Because they are in series, the net capacitance is calculated as:

$C=\frac{{C}_{1}\times {C}_{2}}{{C}_{1}+{C}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{2{K}_{1}{\in}_{0}A}{d}}\times {\displaystyle \frac{2{K}_{2}{\in}_{0}A}{d}}}{{\displaystyle \frac{2{K}_{1}{\in}_{0}A}{d}}+{\displaystyle \frac{2{K}_{2}{\in}_{0}A}{d}}}\phantom{\rule{0ex}{0ex}}=\frac{2{K}_{1}{K}_{2}{\in}_{0}A}{d\left({K}_{1}+{K}_{2}\right)}$

(b) Here, the capacitor has three parts. These can be taken in series.

Now,

${C}_{1}=\frac{{K}_{1}{\in}_{0}A}{\left({\displaystyle \frac{d}{3}}\right)}=\frac{3{K}_{1}{\in}_{0}A}{d}$

${C}_{2}\mathit{=}\frac{\mathit{3}\mathit{}{\mathit{K}}_{\mathit{2}}\mathit{}{\mathit{\in}}_{\mathit{0}}\mathit{}\mathit{A}}{\mathit{d}}\phantom{\rule{0ex}{0ex}}{C}_{\mathit{3}}\mathit{=}\frac{\mathit{3}\mathit{}{\mathit{K}}_{\mathit{3}}\mathit{}{\mathit{\in}}_{\mathit{0}}\mathit{}\mathit{A}}{\mathit{d}}\phantom{\rule{0ex}{0ex}}$

Thus, the net capacitance is calculated as:

$C=\frac{{C}_{1}\times {C}_{2}\times {C}_{2}}{{C}_{1}{C}_{2}+{C}_{2}{C}_{3}+{C}_{3}{C}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{3{K}_{1}{\in}_{0}A}{d}}\times {\displaystyle \frac{3{K}_{2}{\in}_{0}A}{d}}\times {\displaystyle \frac{3{K}_{3}{\in}_{0}A}{d}}}{{\displaystyle \frac{3{K}_{1}{\in}_{0}A}{d}}\times {\displaystyle \frac{3{K}_{2}{\in}_{0}A}{d}}+{\displaystyle \frac{3{K}_{2}{\in}_{0}A}{d}}\times {\displaystyle \frac{3{K}_{3}{\in}_{0}A}{d}}+{\displaystyle \frac{3{K}_{3}{\in}_{0}A}{d}}\times {\displaystyle \frac{3{K}_{1}{\in}_{0}A}{d}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{3{\in}_{0}{K}_{1}{K}_{2}{K}_{3}}{d({K}_{1}{K}_{2}+{K}_{2}{K}_{3}+{K}_{3}{K}_{1})}$

(c)

Here,

${C}_{1}=\frac{{K}_{1}{\in}_{0}A/2}{d}=\frac{{K}_{1}{\in}_{0}A}{2d}\phantom{\rule{0ex}{0ex}}{C}_{2}=\frac{{K}_{2}{\in}_{0}A}{2d}$

These two parts are in parallel.

$\therefore C\mathit{=}{C}_{\mathit{1}}\mathit{+}{C}_{2}\phantom{\rule{0ex}{0ex}}=\frac{{\in}_{0}\mathrm{A}}{2d}\left({K}_{1}+{K}_{2}\right)$

#### Page No 169:

#### Question 57:

A capacitor is formed by two square metal-plates of edge *a*, separated by a distance *d*. Dielectrics of dielectric constant *K*_{1} and *K*_{2}_{ }are filled in the gap as shown in figure (31-E25). Find the capacitance.

Figure

#### Answer:

Let us consider an elemental capacitor of width *dx* at a distance *x* from the left end of the capacitor. It has two capacitive elements of dielectric constants *K*_{1} and *K*_{2}_{ }with plate separations (*x* tan *θ*) and (*d* − *x* tan *θ*) in series, respectively. The areas of the plates of the capacitors are *adx*.

The capacitances of the capacitive elements of the elemental capacitor are:

$d{C}_{1}=\frac{{\in}_{0}{K}_{2}\left(adx\right)}{x\mathrm{tan}\theta},d{C}_{2}=\frac{{\in}_{0}{K}_{1}\left(adx\right)}{d-x\mathrm{tan}\theta}$

The net capacitance of the elemental capacitor is given by

$\frac{1}{dC}=\frac{1}{d{C}_{1}}+\frac{1}{d{C}_{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{dC}=\frac{x\mathrm{tan}\theta}{{\in}_{0}{K}_{2}\left(a\mathrm{d}x\right)}+\frac{d-x\mathrm{tan}\theta}{{\in}_{0}{K}_{1}\left(adx\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow dC=\frac{{\in}_{0}{K}_{1}{K}_{2}\left(adx\right)}{{K}_{1}x\mathrm{tan}\theta +{K}_{2}(d-x\mathrm{tan}\theta )}\phantom{\rule{0ex}{0ex}}$

Thus, integrating the above expression to calculate the net capacitance

$\mathrm{C}={\int}_{0}^{\mathrm{a}}dC={\int}_{0}^{a}\frac{{\in}_{0}{K}_{1}{K}_{2}adx}{{K}_{1}x\mathrm{tan}\theta +{K}_{2}(d-x\mathrm{tan}\theta )}\phantom{\rule{0ex}{0ex}}C={\in}_{0}{K}_{1}{K}_{2}a{\int}_{0}^{a}\frac{dx}{{K}_{2}d+x\mathrm{tan}\theta ({K}_{1}-{K}_{2})}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\in}_{0}{K}_{1}{K}_{2}a}{\mathrm{tan}\theta ({K}_{1}-{K}_{2})}{\left[{\mathrm{log}}_{\mathrm{e}}[{K}_{2}d+x\mathrm{tan}\theta ({K}_{1}-{K}_{2}\left)\right]\right]}_{0}^{a}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\in}_{0}{K}_{1}{K}_{2}a}{\mathrm{tan}\theta ({K}_{1}-{K}_{2})}\left[{\mathrm{log}}_{\mathrm{e}}[{K}_{2}d+a\mathrm{tan}\theta ({K}_{1}-{K}_{2}\left)\right]-{\mathrm{log}}_{\mathrm{e}}{K}_{2}d\right]$

As we know that $\mathrm{tan}\theta =\frac{d}{a}$ substituting in the expression for capacitance *C*.

Now,

$\Rightarrow C=\frac{{\in}_{0}{K}_{1}{K}_{2}a}{{\displaystyle \frac{d}{a}}\times ({K}_{1}-{K}_{2})}\left[{\mathrm{log}}_{\mathrm{e}}[{K}_{2}d+a\times \frac{d}{a}({K}_{1}-{K}_{2}\left)\right]-{\mathrm{log}}_{\mathrm{e}}{K}_{2}d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\in}_{0}{K}_{1}{K}_{2}a}{{\displaystyle \frac{d}{a}}\times ({K}_{1}-{K}_{2})}\left[{\mathrm{log}}_{\mathrm{e}}{K}_{1}d-{\mathrm{log}}_{\mathrm{e}}{K}_{2}d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\in}_{0}{K}_{1}{K}_{2}{a}^{2}}{d({K}_{1}-{K}_{2})}\left[{\mathrm{log}}_{\mathrm{e}}\left(\frac{{K}_{1}}{{K}_{2}}\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 169:

#### Question 58:

Figure (31-E26) shows two identical parallel plate capacitors connected to a battery through a switch *S*. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

Figure

#### Answer:

When the switch is closed, both capacitors are in parallel.

$\Rightarrow $The total energy of the capacitor when the switch is closed is given by

*E _{i}*$=\frac{1}{2}C{V}^{2}+\frac{1}{2}C{V}^{2}=C{V}^{2}$

When the switch is opened and the dielectric is induced, the capacitance of the capacitor A becomes

$C\text{'}=KC=3C$

The energy stored in the capacitor A is given by

${E}_{\mathrm{A}}=\frac{1}{2}C\text{'}{V}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{\mathrm{A}}=\frac{1}{2}\left(3C\right){V}^{2}=\frac{3}{2}C{V}^{2}$

The energy in the capacitor B is given by

${E}_{\mathrm{B}}=\frac{1}{2}\times \frac{C}{3}\times {V}^{2}$

∴ Total final energy

${E}_{\mathrm{f}}={E}_{\mathrm{A}}+{E}_{\mathrm{B}}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{\mathrm{f}}=\frac{3}{2}C{V}^{2}+\frac{1}{6}C{V}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{\mathrm{f}}=\frac{9C{V}^{2}+1C{V}^{2}}{6}=\frac{10}{6}C{V}^{2}$

Now,

Ratio of the energies, $\frac{{E}_{1}}{{E}_{2}}=\frac{C{V}^{2}}{{\displaystyle \frac{10}{6}C{V}^{2}}}=\frac{3}{5}$

#### Page No 169:

#### Question 59:

A parallel-plate capacitor of plate area *A* and plate separation *d* is charged to a potential difference *V* and then the battery is disconnected. A slab of dielectric constant *K* is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.

#### Answer:

Initial capacitance, $C=\frac{{\in}_{0}A}{d}$

The energy of the capacitor before the insertion of the dielectric is given by

${E}_{1}=\frac{1}{2}C{V}^{2}=\frac{1}{2}\frac{{\in}_{0}A}{d}{V}^{2}$

After inserting the dielectric slab, the capacitance becomes

${C}_{1}=KC=K\frac{{\in}_{0}A}{d}$

and the final voltage becomes

${V}_{1}=\frac{V}{K}$

Thus, the final energy stored in the capacitor is given by

${E}_{2}=\frac{1}{2}\frac{K{\in}_{0}A}{d}\times {\left(\frac{V}{K}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {E}_{2}=\frac{1}{2}\frac{{\in}_{0}\mathit{}A{V}^{2}}{Kd}$

Now,

Work done = Change in energy

∴ *W* = *E*_{2} $-$ *E*_{1}

$\Rightarrow W=\frac{1}{2}\frac{{\in}_{0}A{V}^{2}}{Kd}-\frac{1}{2}\frac{{\in}_{0}A{V}^{\mathit{2}}}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow W=\frac{1}{2}\frac{{\in}_{0}A{V}^{2}}{d}\left(\frac{1}{K}-1\right)$

#### Page No 169:

#### Question 60:

A capacitor having a capacitance of 100 µF is charged to a potential difference of 50 V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2⋅5 is inserted. Calculate the new potential difference between the plates. (c) What charge would have produced this potential difference in absence of the dielectric slab. (d) Find the charge induced at a surface of the dielectric slab.

#### Answer:

(a) The magnitude of the charge can be calculated as:

Charge = Capacitance × Potential difference

$\Rightarrow Q=100\times {10}^{-6}\times 50=5\mathrm{mC}$

(b) When a dielectric is introduced, the potential difference decreases.

We know,

$V=\frac{\mathrm{Initial}\mathrm{potential}}{\mathrm{Dielectric}\mathrm{constant}}\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{50}{2.5}=20\mathrm{V}$

(c) Now, the charge on the capacitance can be calculated as:

Charge = Capacitance × Potential difference

$\Rightarrow {q}_{f}=20\times 100\times {10}^{-6}=2\mathrm{mC}$

(d) The charge induced on the dielectric can be calculated as:

$q={q}_{i}\left(1-\frac{1}{K}\right)=5\mathrm{mC}\left(1-\frac{1}{2.5}\right)=3\mathrm{mC}$

#### Page No 169:

#### Question 61:

A sphercial capacitor is made of two conducting spherical shells of radii *a* and *b*. The space between the shells is filled with a dielectric of dielectric constant *K* up to a radius *c* as shown in figure (31-E27). Calculate the capacitance.

Figure

#### Answer:

We have two capacitors: one made by the shells a and c and the other made by the shells b and c.

The capacitance of the capacitor *C*_{ac} is given by

${C}_{\mathrm{ac}}=\frac{4\pi {\in}_{0}acK}{(c-a)}$

The capacitance of the capacitor *C*_{cb} is given by

${C}_{\mathrm{cb}}=\frac{4\pi {\in}_{0}bcK}{K(b-c)}$

The two capacitors are in series; thus, the equivalent capacitance is given by

$\frac{1}{C}=\frac{1}{{C}_{\mathrm{ac}}}+\frac{1}{{C}_{\mathrm{cb}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{C}=\frac{(c-a)}{4\pi {\in}_{0}acK}+\frac{(b-c)}{4\pi {\in}_{0}cb}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{C}=\frac{b(c-a)+Ka(b-c)}{K4\pi {\in}_{0}abc}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{K4\pi {\in}_{0}abc}{b(c-a)+Ka(b-c)}$

#### Page No 170:

#### Question 62:

Consider an assembly of three conducting concentric spherical shell of radii *a*, *b* and *c* as shown in figure (31-E28). Find the capacitance of the assembly between the points *A* and *B*.

#### Answer:

The spherical shells form two spherical capacitors: one made by A and B and the other made by B and C.

The capacitance of the spherical capacitor made by the shells of radii *r*_{1} and *r*_{2} is given by

$C=\frac{4\pi {\in}_{0}}{\left[{\displaystyle \frac{1}{{r}_{1}}}-{\displaystyle \frac{1}{{r}_{2}}}\right]}=\frac{4\pi {\in}_{0}{r}_{1}{r}_{2}}{{r}_{2}-{r}_{1}}$

The capacitance of the capacitor made by A and B is given by

${C}_{\mathrm{AB}}=\frac{4\pi {\in}_{0}ab}{b-a}$

The capacitance of the capacitor made by B and C is given by

${C}_{\mathrm{BC}}=\frac{4\pi {\in}_{0}bc}{c-b}$

As the capacitors are in series, the net capacitance is given by

$\frac{1}{C}=\frac{1}{{C}_{\mathrm{AB}}}+\frac{1}{{C}_{\mathrm{BC}}}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{C}_{\mathrm{AB}}{C}_{\mathrm{BC}}}{{C}_{\mathrm{AB}}+{C}_{\mathrm{BC}}}=\frac{{\displaystyle \frac{(4\pi {\in}_{0}{)}^{2}a{b}^{2}c}{(b-a)(c-b)}}}{{\displaystyle \frac{4\pi {\in}_{0}ab}{(b-a)}}+{\displaystyle \frac{4\pi {\in}_{0}bc}{(c-b)}}}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\displaystyle \frac{4\pi {\in}_{0}a{b}^{2}c}{(b-a)(c-b)}}}{{\displaystyle \left(\frac{ab(c-b)+bc(b-a)}{(b-a)(c-b)}\right)}}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{4\pi {\in}_{0}a{b}^{2}c}{\left[ab\right(c-b)+bc(b-a\left)\right]}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{4\pi {\in}_{0}a{b}^{2}c}{{b}^{2}(c-a)}=\frac{4\pi {\in}_{0}ac}{(c-a)}$

#### Page No 170:

#### Question 63:

Suppose the space between the two inner shells of the previous problem is filled with a dielectric of dielectric constant *K*. Find the capacitance of the system between *A* and *B*.

Figure

#### Answer:

Since the space between the two inner shells is filled with a dielectric, capacitance *C*_{AB} becomes ${C}_{\mathrm{AB}}=\frac{4\pi {\in}_{0}abK}{(b-a)}$ and capacitance *C*_{BC}_{ }becomes ${C}_{\mathrm{BC}}=\frac{4\pi {\in}_{0}bc}{(c-b)}$.

Now, as the capacitors are in series, the equivalent capacitance is given by

$\frac{1}{C}=\frac{1}{{C}_{\mathrm{AB}}}+\frac{1}{{C}_{\mathrm{BC}}}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{C}_{\mathrm{AB}}{C}_{\mathrm{BC}}}{{C}_{\mathrm{AB}}+{C}_{\mathrm{BC}}}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\displaystyle \frac{(4\pi {\in}_{0}{)}^{2}a{b}^{2}Kc}{(b-a)(c-b)}}}{{\displaystyle \frac{4\pi {\in}_{0}abk}{(b-a)}}+{\displaystyle \frac{4\pi {\in}_{0}bc}{(c-b)}}}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\displaystyle \frac{4\pi {\in}_{0}Ka{b}^{2}c}{(b-a)(c-b)}}}{{\displaystyle \left(\frac{abk(c-b)+bc(b-a)}{(b-a)(c-b)}\right)}}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{4\pi {\in}_{0}Ka{b}^{2}c}{\left[abk\right(c-b)+bc(b-a\left)\right]}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{4\pi {\in}_{0}Kabc}{\left[ak\right(c-b)+c(b-a\left)\right]}$

#### Page No 170:

#### Question 64:

An air-filled parallel-plate capacitor is to be constructed which can store 12 µC of charge when operated at 1200 V. What can be the minimum plate area of the capacitor? The dielectric strength of air is $3\times {10}^{6}\mathrm{V}{\mathrm{m}}^{-1}.$

#### Answer:

Charge that the capacitor can hold = 12 μC

Operating voltage = 1200 V

Breakdown strength, *b* = $3\times {10}^{6}\mathrm{V}{\mathrm{m}}^{-1}$

The separation between the plates of the capacitor is given by

$d=\frac{V}{b}=\frac{1200}{3\times {10}^{6}}=4\times {10}^{-4}\mathrm{m}$

The capacitance of the capacitor is given by

$C=\frac{Q}{V}=\frac{12\times {10}^{-6}}{1200}={10}^{-8}\mathrm{F}$

The capacitance can be expressed in terms of the area of the plates (*A*) and the separation of the plates (*d*) as:

$C=\frac{{\in}_{0}A}{d}={10}^{-8}\mathrm{F}\phantom{\rule{0ex}{0ex}}$

Thus, the area of the plates is given by

$A=\frac{{10}^{-8}\times (4\times {10}^{-4})}{8.854\times {10}^{-12}}=0.45{\mathrm{m}}^{2}$

#### Page No 170:

#### Question 65:

A parallel-plate capacitor with the plate area 100 cm^{2} and the separation between the plates 1⋅0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.

#### Answer:

Area of the plates of the capacitor, *A* = 100 cm^{2} = 10^{$-$2} m^{2}

Separation between the plates, *d* = 1 cm = 10^{$-$2} m

Emf of battery, *V* = 24 V

Therefore,

Capacitance, $C=\frac{{\in}_{0}A}{d}=\frac{(8.85\times {10}^{-12})\times \left({10}^{-2}\right)}{{10}^{-2}}=8.85\times {10}^{-12}\mathrm{V}$

Energy stored in the capacitor, $E=\frac{1}{2}C{V}^{2}=\frac{1}{2}\times (8.85\times {10}^{-12})\times (24{)}^{2}\phantom{\rule{0ex}{0ex}}=2548.8\times {10}^{-12}\mathrm{J}$

Force of attraction between the plates, $F=\frac{E}{d}=\frac{2548.8\times {10}^{-12}}{{10}^{-2}}=2548.8\times {10}^{-10}\mathrm{N}$

#### Page No 170:

#### Question 66:

Consider the situation shown in figure (31-E29). The width of each plate is *b*. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf ε. All surfaces are frictionless. Calculate the value of *M* for which the dielectric slab will stay in equilibrium.

Figure

#### Answer:

Let the potential of the battery connected to the capacitor be *V*.

Let the length of the part of the slab inside the capacitor be *x*.

The capacitor can be considered to be two capacitors in parallel.

The capacitances of the two capacitors in parallel are:

${C}_{1}=\frac{{K}_{1}{\in}_{0}bx}{d},{C}_{2}=\frac{{\in}_{0}b({l}_{1}-x)}{d}$

*C*_{1}_{ }is the part of the capacitor having the dielectric inserted in it and *C*_{2} is the capacitance of the part of the capacitor without dielectric.

As, *C*_{1} and *C*_{2} are in parallel.

Therefore, the net capacitance is given by

C = *C*_{1} + *C*_{2}

$C=\frac{{K}_{1}{\in}_{0}bx}{d}+\frac{{\in}_{0}b({l}_{1}-x)}{d}\phantom{\rule{0ex}{0ex}}C=\frac{{\in}_{0}b}{d}\left[{l}_{1}+x({K}_{1}-1)\right]$

The dielectric slab is attracted by the electric field of the capacitor and applies a force in left direction.

Let us consider electric force of magnitude *F* pulls the slab in left direction.

Let there be an infinitesimal displacement *dx** *in left direction by the force *F*.

The work done by the force = *F*.*dx*

Let the potential of the battery connected to the left capacitor be *V*_{1} and that of the battery connected with the right capacitor be *V*_{2}_{.}

With the displacement of slab, the capacitance will increase, therefore the energy of the capacitor will also increase. In order to maintain constant voltage, the battery will supply extra charges, therefore the battery will do work.

Work done by the battery = change in energy of capacitor + work done by the force *F* on the capacitor

*dW*_{B}_{ }= *dU*_{ }+ *dW*_{F}

Let the charge d*q* is supplied by the battery, and the change in the capacitor be* dC*

*dW*_{B} = (*dq*)*.**V* = (*dC*)*.V*^{2}

*dU* = $\frac{1}{2}$(*dC*)*.V*^{2}

(*dC*)*.V*^{2} = $\frac{1}{2}$(*dC*)*.V*^{2} + *F*.*dx*

$\frac{1}{2}$(*dC*)*.V*^{2}* = F.dx*

$\Rightarrow F=\frac{1}{2}\frac{\mathit{d}\mathit{C}}{\mathit{d}\mathit{x}}{V}_{1}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{1}{2}\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left(\frac{{\in}_{0}b}{d}\left[{l}_{1}+x({K}_{1}-1)\right]\right){V}_{1}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{{\in}_{0}b{V}_{1}^{2}({K}_{1}-1)}{2d}$

To keep the dielectric slab in equilibrium, the electrostatic force on it must be counteracted by the weight of the block attached.

Therefore,

$\frac{{\in}_{0}b{V}^{2}}{2d}\left(K-1\right)=Mg\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{{\in}_{0}b{V}^{2}}{2dg}\left(K-1\right)$

#### Page No 170:

#### Question 67:

Figure (31-E30) shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width *b* and lengths *l*_{1} and *l*_{2}. The left half of the dielectric slab has a dielectric constant *K*_{1} and the right half *K*_{2}. Neglecting any friction, find the ration of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium.

#### Answer:

Let the potential of the battery connected to the left capacitor be *V*_{1} and that of the battery connected to the right capacitor be *V*_{2}_{.}

Considering the left capacitor,

Let the length of the part of the slab inside the capacitor be *x*.

The left capacitor can be considered to be two capacitors in parallel.

The capacitances of the two capacitors in parallel are:

${C}_{1}=\frac{{K}_{1}{\in}_{0}bx}{d},{C}_{2}=\frac{{\in}_{0}b({l}_{1}-x)}{d}$

*C*_{1}_{ }is the part of the capacitor having the dielectric inserted in it and *C*_{2} is the capacitance of the part of the capacitor without dielectric.

As, *C*_{1} and *C*_{2} are in parallel.

Therefore, the net capacitance is given by

C = *C*_{1} + *C*_{2}

$C=\frac{{K}_{1}{\in}_{0}bx}{d}+\frac{{\in}_{0}b({l}_{1}-x)}{d}\phantom{\rule{0ex}{0ex}}C=\frac{{\in}_{0}b}{d}\left[{l}_{1}+x({K}_{1}-1)\right]$

Therefore, the potential energy stored in the left capacitor will be

$U=\frac{1}{2}C{V}_{1}^{2}\phantom{\rule{0ex}{0ex}}U=\frac{{\in}_{0}b{V}_{1}^{2}}{2d}\left[{l}_{1}+x({K}_{1}-1)\right]...\left(1\right)$

The dielectric slab is attracted by the electric field of the capacitor and applies a force in left direction.

Let us consider electric force of magnitude *F* pulls the slab in left direction.

Let there be an infinitesimal displacement d*x** *in left direction by the force *F*.

The work done by the force = *F*.d*x*

Let us consider a small displacement *dx* of the slab in the inward direction. The capacitance will increase, therefore the energy of the capacitor will also increase. In order to maintain constant voltage, the battery will supply extra charges, therefore the battery will do work.

Work done by the battery = change in energy of capacitor + work done by the force *F* on the capacitor

*dW*_{B}_{ }= *dU*_{ }+ *dW*_{F}

Let the charge d*q* is supplied by the battery, and the change in the capacitor be d*C*

*dW*_{B} = (*dq*)*.**V* = (*dC*)*.V*^{2}

*dU* = $\frac{1}{2}$(*dC*)*.V*^{2}

(*dC*)*.V*^{2} = $\frac{1}{2}$(*dC*)*.V*^{2} + *F*.*dx*

$\frac{1}{2}$(*dC*)*.V*^{2}* = F.dx*

$\Rightarrow F=\frac{1}{2}\frac{dC}{dx}{V}_{1}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{1}{2}\frac{d}{dx}\left(\frac{{\in}_{0}b}{d}\left[{l}_{1}+x({K}_{1}-1)\right]\right){V}_{1}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{{\in}_{0}b{V}_{1}^{2}({K}_{1}-1)}{2d}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{1}^{2}=\frac{F\times 2d}{{\in}_{0}b({K}_{1}-1)}\Rightarrow {V}_{1}=\sqrt{\frac{F\times 2d}{{\in}_{0}b({K}_{1}-1)}}$

Similarly, for the right side the voltage of the battery is given by

${V}_{2}=\sqrt{\frac{F\times 2d}{{\in}_{0}b({K}_{2}-1)}}$

Thus, the ratio of the voltages is given by

$\frac{{V}_{1}}{{V}_{2}}=\frac{\sqrt{\frac{F\times 2d}{{\in}_{0}b({\mathrm{K}}_{1}-1)}}}{\sqrt{\frac{F\times 2d}{{\in}_{0}b({\mathrm{K}}_{2}-1)}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{V}_{1}}{{V}_{2}}=\frac{\sqrt{{K}_{2}-1}}{\sqrt{{K}_{1}-1}}$

Thus, the ratio of the emfs of the left battery to the right battery is $\frac{{V}_{1}}{{V}_{2}}=\frac{\sqrt{{K}_{2}-1}}{\sqrt{{K}_{1}-1}}$

#### Page No 170:

#### Question 68:

Consider the situation shown in figure (31-E31). The plates of the capacitor have plate area *A* and are clamped in the laboratory. The dielectric slab is released from rest with a length *a* inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.

Figure

#### Answer:

Given that the area of the plates of the capacitors is *A*.

As ''*a*'' length of the dielecric slab is inside the capacitor.

Therefore, the area of the plate covered with dielectric is = $\frac{A}{l}a$

The capacitance of the portion with dielectric is given by

${C}_{1}=\frac{K{\in}_{0}Aa}{ld}$

The capacitance of the portion without dielectric is given by

${C}_{1}=\frac{{\in}_{0}A(l-a)}{ld}$

The two parts can be considered to be in parallel.

Therefore, the net capacitance is given by

$C={C}_{1}+{C}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\in}_{0}A}{ld}\left[Ka+(l-a)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow C=\frac{{\in}_{0}A}{ld}\left[l+a(K-1)\right]$

Let us consider a small displacement *da* of the slab in the inward direction. The capacitance will increase, therefore the energy of the capacitor will also increase. In order to maintain constant voltage, the battery will supply extra charges, therefore the battery will do work.

Work done by the battery = change in energy of capacitor + work done by the force *F* on the capacitor

*dW*_{B}_{ }= *dU*_{ }+ *dW*_{F}

Let the charge* dq* is supplied by the battery, and the change in the capacitor be* dC*

*dW*_{B} = (*dq*)*.**V* = (*dC*)*.V*^{2}

*dU* = $\frac{1}{2}$(*dC*)*.V*^{2}

(d*C*)*.V*^{2} = $\frac{1}{2}$(d*C*)*.V*^{2} + *F*.d*a*

$\frac{1}{2}$(d*C*)*.V*^{2}* = F*.d*a*

$\Rightarrow F=\frac{1}{2}\frac{\mathrm{d}C}{\mathrm{d}a}{V}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}a}\left(\frac{{\in}_{0}A}{ld}\left[l+a(K-1)\right]\right){V}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{1}{2}\frac{{\in}_{0}A}{ld}(K-1)$

The acceleration of the dielectric is given by *a*_{0}= $\frac{1}{2}\frac{{\in}_{0}A}{ldm}(K-1)$

As, the force is in inward direction, it will tend to make the dielectric to completely fill the space inside the capacitors. As, the dielectric completely fills the space inside the capacitor at this instant its velocity is not zero. The dielectric slab tends to move outside the capacitor. As the slab tends to move out, the direction of the force due to the capacitor will reverse its direction. Thus, the dielectric slab will have a periodic motion.

The time taken to move distance (*l$-$a*) can be calculated as:

$(l-a)=\frac{1}{2}{a}_{0}{t}^{2}\phantom{\rule{0ex}{0ex}}t=\sqrt{\frac{2\left(l-a\right)}{{a}_{0}}}\phantom{\rule{0ex}{0ex}}t=\sqrt{2\left(l-a\right)\times \frac{2ldm}{{\in}_{0}A{V}^{2}(K-1)}}\phantom{\rule{0ex}{0ex}}t=\sqrt{\frac{4m\left(l-a\right)ld}{{\in}_{0}A{V}^{2}(K-1)}}$

For the complete cycle the time period will be four times the time taken for covering distance (*l$-$a*).

It is given by :

$T=4t=4\times 2\sqrt{\frac{m\left(l-a\right)ld}{{\in}_{0}A{V}^{2}(K-1)}}=8\sqrt{\frac{m\left(l-a\right)ld}{{\in}_{0}A{V}^{2}(K-1)}}$

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