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#### Page No 87:

#### Question 2.1:

Two charges 5 ×
10^{−8} C and −3 × 10^{−8} C
are located 16 cm apart. At what point(s) on the line joining the two
charges is the electric potential zero? Take the potential at
infinity to be zero.

#### Answer:

There are two charges,

Distance between the
two charges, *d* = 16 cm = 0.16 m

Consider a point P on the line joining the two charges, as shown in the given figure.

*r* = Distance of
point P from charge *q*_{1}

Let the electric
potential (*V*) at point P be zero.

Potential at point P is
the sum of potentials caused by charges *q*_{1} and *q*_{2}
respectively.

Where,

= Permittivity of free space

For *V* = 0,
equation (i) reduces to

Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

Suppose point P is
outside the system of two charges at a distance *s *from the
negative charge, where potential is zero, as shown in the following
figure.

For this arrangement, potential is given by,

For *V* = 0,
equation (ii) reduces to

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

#### Page No 87:

#### Question 2.2:

A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.

#### Answer:

The given figure shows
six equal amount of charges, *q*, at the vertices of a regular
hexagon.

Where,

Charge, *q* = 5 µC
= 5 × 10^{−6} C

Side of the hexagon, *l*
= AB = BC = CD = DE = EF = FA = 10 cm

Distance of each vertex
from centre O, *d* = 10 cm

Electric potential at point O,

Where,

= Permittivity of free space

Therefore, the
potential at the centre of the hexagon is 2.7 ×
10^{6} V.

#### Page No 87:

#### Question 2.3:

Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.

**(a)** Identify an
equipotential surface of the system.

**(b)** What is the
direction of the electric field at every point on this surface?

#### Answer:

**(a) **The
situation is represented in the given figure.

An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.

**(b) **The direction of the electric field at every point on this
surface is normal to the plane in the direction of AB.

#### Page No 87:

#### Question 2.4:

A spherical conductor
of radius 12 cm has a charge of 1.6 × 10^{−7}C
distributed uniformly on its surface. What is the electric field

**(a) ** Inside the
sphere

**(b) ** Just
outside the sphere

**(c) ** At a point
18 cm from the centre of the sphere?

#### Answer:

**(a) **Radius of
the spherical conductor, *r* = 12 cm = 0.12 m

Charge
is uniformly distributed over the conductor, *q *= 1.6 ×
10^{−7} C

Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.

**(b) **Electric
field *E* just outside the conductor is given by the relation,

Where,

= Permittivity of free space

Therefore, the electric field just outside the sphere is .

**(c) **Electric
field at a point 18 m from the centre of the sphere = *E*_{1}

Distance
of the point from the centre, *d* = 18 cm = 0.18 m

Therefore, the electric field at a point 18 cm from the centre of the sphere is

.

#### Page No 87:

#### Question 2.5:

A parallel plate
capacitor with air between the plates has a capacitance of 8 pF (1pF
= 10^{−12} F). What will be the capacitance if the
distance between the plates is reduced by half, and the space between
them is filled with a substance of dielectric constant 6?

#### Answer:

Capacitance between the parallel plates of the capacitor, C = 8 pF

Initially, distance
between the parallel plates was *d *and it was filled with air.
Dielectric constant of air, *k* = 1

Capacitance, *C*,
is given by the formula,

Where,

*A* = Area of each
plate

= Permittivity of free space

If distance between the
plates is reduced to half, then new distance, *d*^{’}
=

Dielectric constant of the substance filled in between the plates, = 6

Hence, capacitance of the capacitor becomes

Taking ratios of equations (i) and (ii), we obtain

Therefore, the capacitance between the plates is 96 pF.

#### Page No 87:

#### Question 2.6:

Three capacitors each of capacitance 9 pF are connected in series.

**(a)** What is the total capacitance of the combination?

**(b)** What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

#### Answer:

**(a) **Capacitance of each of the three capacitors, *C* = 9 pF

Equivalent capacitance (*C*^{’}) of the combination of the capacitors is given by the relation,

$\frac{1}{C\text{'}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{C\text{'}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow C\text{'}=3\mathrm{pF}$

Therefore, total capacitance of the combination is $3\mathrm{pF}$.

**(b) **Supply voltage, *V* = 120 V

Potential difference (*V*') across each capacitor is equal to one-third of the supply voltage.

Therefore, the potential difference across each capacitor is 40 V.

#### Page No 87:

#### Question 2.7:

Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

**(a)**
What is the total capacitance of the combination?

**(b)**
Determine the charge on each capacitor if the combination is
connected to a 100 V supply.

#### Answer:

**(a) **Capacitances
of the given capacitors are

For the parallel combination of the capacitors, equivalent capacitoris given by the algebraic sum,

Therefore, total capacitance of the combination is 9 pF.

**(b) **Supply
voltage, *V*
= 100 V

The voltage through all the three capacitors is
same = *V*
= 100 V

Charge
on a capacitor of capacitance *C*
and potential difference *V*
is given by the relation,

*q*
= *VC* …
(i)

For C = 2 pF,

For C = 3 pF,

For C = 4 pF,

#### Page No 87:

#### Question 2.8:

In
a parallel plate capacitor with air between the plates, each plate
has an area of 6 × 10^{−3}
m^{2}
and the distance between the plates is 3 mm. Calculate the
capacitance of the capacitor. If this capacitor is connected to a 100
V supply, what is the charge on each plate of the capacitor?

#### Answer:

Area
of each plate of the parallel plate capacitor, *A*
= 6 ×
10^{−3}
m^{2}

Distance between the
plates, *d* = 3 mm = 3 ×
10^{−3} m

Supply voltage, *V*
= 100 V

Capacitance *C* of
a parallel plate capacitor is given by,

Where,

= Permittivity of free space

= 8.854 ×
10^{−12} N^{−1} m^{−2} C^{−2}

Therefore, capacitance
of the capacitor is 17.71 pF and charge on each plate is 1.771 ×
10^{−9} C.

#### Page No 88:

#### Question 2.9:

Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

**(a)
** While the voltage supply remained
connected.

**(b)
** After the supply was disconnected.

#### Answer:

**(a)** Dielectric
constant of the mica sheet, *k* = 6

Initial
capacitance, *C* = 1.771 ×
10^{−11} F

Supply
voltage, *V *= 100 V

Potential across the plates remains 100 V.

**(b)** Dielectric
constant,* k* = 6

Initial
capacitance, *C* = 1.771 ×
10^{−11} F

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge
= 1.771 × 10^{−9}
C

Potential across the plates is given by,

#### Page No 88:

#### Question 2.10:

A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

#### Answer:

Capacitor of the
capacitance, *C* = 12 pF = 12 ×
10^{−12} F

Potential difference, *V
*= 50 V

Electrostatic energy stored in the capacitor is given by the relation,

Therefore, the electrostatic energy stored in the capacitor is

#### Page No 88:

#### Question 2.11:

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

#### Answer:

Capacitance
of the capacitor, *C*
= 600 pF

Potential
difference, *V*
= 200 V

Electrostatic energy stored in the capacitor is given by,

If
supply is disconnected from the capacitor and another capacitor of
capacitance *C*
= 600 pF is connected to it, then equivalent capacitance (*C*^{’})
of the combination is given by,

New electrostatic energy can be calculated as

Therefore, the electrostatic energy lost in the process is.

#### Page No 88:

#### Question 2.12:

A charge of 8 mC is
located at the origin. Calculate the work done in taking a small
charge of −2 × 10^{−9} C from a point P (0,
0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

#### Answer:

Charge located at the
origin, *q* = 8 mC= 8 × 10^{−3 }C

Magnitude of a small
charge, which is taken from a point P to point R to point Q, *q*_{1}
= − 2 × 10^{−9 }C

All the points are represented in the given figure.

Point P is at a
distance, *d*_{1} = 3 cm, from the origin along *z*-axis.

Point Q is at a
distance, *d*_{2} = 4 cm, from the origin along *y*-axis.

Potential at point P,

Potential at point Q,

Work done (*W*) by
the electrostatic force is independent of the path.

Therefore, work done during the process is 1.27 J.

#### Page No 88:

#### Question 2.13:

A cube of side *b *has
a charge *q *at each of its vertices. Determine the potential
and electric field due to this charge array at the centre of the
cube.

#### Answer:

Length of the side of a
cube = *b*

Charge at each of its
vertices = *q*

A cube of side *b *is
shown in the following figure.

*d* = Diagonal of
one of the six faces of the cube

*l* = Length of
the diagonal of the cube

The electric potential
(*V*) at the centre of the cube is due to the presence of eight
charges at the vertices.

Therefore, the potential at the centre of the cube is .

The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the centre of the cube. Hence, the electric field is zero at the centre.

#### Page No 88:

#### Question 2.14:

Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

**(a)** at the
mid-point of the line joining the two charges, and

**(b)** at a point 10 cm from this midpoint in a plane normal to
the line and passing through the mid-point.

#### Answer:

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

Magnitude of charge
located at A, *q*_{1} = 1.5 μC

Magnitude of charge
located at B, *q*_{2} = 2.5 μC

Distance between the
two charges, *d* = 30 cm = 0.3 m

**(a) **Let *V*_{1}
and *E*_{1} are the electric potential and electric
field respectively at O.

*V*_{1}
= Potential due to charge at A + Potential due to charge at B

Where,

∈_{0}
= Permittivity of free space

*E*_{1}
= Electric field due to *q*_{2} − Electric field
due to *q*_{1}

Therefore,
the potential at mid-point is 2.4 × 10^{5} V and the
electric field at mid-point is 4× 10^{5} V m^{−1}.
The field is directed from the larger charge to the smaller charge.

**(b) **Consider a point Z such that normal distanceOZ =
10 cm = 0.1 m, as shown in the following figure.

*V*_{2}
and *E*_{2 }are the electric potential and electric
field respectively at Z.

It can be observed from the figure that distance,

*V*_{2}=
Electric potential due to A + Electric Potential due to B

Electric
field due to *q* at Z,

Electric
field due to *q*_{2} at *Z*,

The resultant field intensity at Z,

Where,
2*θ*is the
angle, ∠*AZ *B

From the figure, we obtain

Therefore,
the potential at a point 10 cm (perpendicular to the mid-point) is
2.0 × 10^{5} V and electric field is 6.6 ×10^{5}
V m^{−1}.

#### Page No 88:

#### Question 2.15:

A spherical conducting shell of inner radius *r*1 and outer
radius *r*2 has a charge *Q*.

**(a) ** A charge *q *is placed at the centre of the shell.
What is the surface charge density on the inner and outer surfaces of
the shell?

**(b) ** Is the electric field inside a cavity (with no charge)
zero, even if the shell is not spherical, but has any irregular
shape? Explain.

#### Answer:

**(a)** Charge placed at the centre of a shell is +*q*.
Hence, a charge of magnitude −*q* will be induced to the
inner surface of the shell. Therefore, total charge on the inner
surface of the shell is −*q*.

Surface charge density at the inner surface of the shell is given by the relation,

A
charge of +*q* is induced on the outer surface of the shell. A
charge of magnitude *Q* is placed on the outer surface of the
shell. Therefore, total charge on the outer surface of the shell is *Q*
+ *q*. Surface charge density at the outer surface of the shell,

**(b) **Yes

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

#### Page No 88:

#### Question 2.16:

**(a)** Show
that the normal component of electrostatic field has a discontinuity
from one side of a charged surface to another given by

Where is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ

**(b)** Show
that the tangential component of electrostatic field is continuous
from one side of a charged surface to another. [Hint: For (a), use
Gauss’s law. For, (b) use the fact that work done by
electrostatic field on a closed loop is zero.]

#### Answer:

**(a) **Electric field on one side of a charged body is *E*_{1}
and electric field on the other side of the same body is *E*_{2}.
If infinite plane charged body has a uniform thickness, then electric
field due to one surface of the charged body is given by,

Where,

= Unit vector normal to the surface at a point

σ = Surface charge density at that point

Electric field due to the other surface of the charged body,

Electric field at any point due to the two surfaces,

Since inside a closed conductor, = 0,

∴

Therefore, the electric field just outside the conductor is .

**(b) **When a
charged particle is moved from one point to the other on a closed
loop, the work done by the electrostatic field is zero. Hence, the
tangential component of electrostatic field is continuous from one
side of a charged surface to the other.

#### Page No 88:

#### Question 2.17:

A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

#### Answer:

Charge
density of the long charged cylinder of length *L*
and radius *r*
is *λ*.

Another
cylinder of same length surrounds the pervious cylinder. The radius
of this cylinder is *R*.

Let
*E* be
the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gauss’s theorem as,

Where,
*d *=
Distance of a point from the common axis of the cylinders

Let
*q* be
the total charge on the cylinder.

It can be written as

Where,

*q*
= Charge on the inner sphere of the outer cylinder

∈_{0}
= Permittivity of free space

Therefore, the electric field in the space between the two cylinders is.

#### Page No 88:

#### Question 2.18:

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

**(a)** Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

**(b)** What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

**(c)** What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

#### Answer:

The distance between electron-proton of a hydrogen atom,

Charge on an electron, *q*_{1}_{ }= −1.6 ×10^{−19}^{ }C

Charge on a proton, *q*_{2}_{ }= +1.6 ×10^{−19}^{ }C

**(a)** Potential at infinity is zero.

Potential energy of the system,* *= Potential energy at infinity − Potential energy at distance *d*

where,

∈_{0} is the permittivity of free space

$\frac{1}{4{\mathrm{\pi \epsilon}}_{0}}=9\times {10}^{9}{\mathrm{Nm}}^{2}{\mathrm{C}}^{-2}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Potential}\mathrm{energy}=0-\frac{9\times {10}^{9}\times {\left(1.6\times {10}^{-19}\right)}^{2}}{0.53\times {10}^{-10}}=-43.47\times {10}^{-19}\mathrm{J}\phantom{\rule{0ex}{0ex}}\because 1.6\times {10}^{-19}\mathrm{J}=1\mathrm{eV}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{Potential}\mathrm{energy}=-43.7\times {10}^{-19}=\frac{-43.7\times {10}^{-19}}{1.6\times {10}^{-19}}=-27.2\mathrm{eV}\phantom{\rule{0ex}{0ex}}$

Therefore, the potential energy of the system is −27.2 eV.

**(b)** Kinetic energy is half of the magnitude of potential energy.

Total energy = 13.6 − 27.2 = 13.6 eV

Therefore, the minimum work required to free the electron is 13.6 eV.

**(c)** When zero of potential energy is taken,

∴Potential energy of the system = Potential energy at *d*_{1} − Potential energy at *d*

#### Page No 89:

#### Question 2.19:

If one of the two
electrons of a H_{2 }molecule is removed, we get a hydrogen
molecular ion.
In the ground state of an,
the two protons are separated by roughly 1.5 Å, and the
electron is roughly 1 Å from each proton. Determine the
potential energy of the system. Specify your choice of the zero of
potential energy.

#### Answer:

The system of two protons and one electron is represented in the given figure.

Charge on proton 1, *q*_{1}
= 1.6 ×10^{−19 }C

Charge on proton 2, *q*_{2}
= 1.6 ×10^{−19 }C

Charge on electron, *q*_{3}
= −1.6 ×10^{−19 }C

Distance between
protons 1 and 2, *d*_{1} = 1.5 ×10^{−10}
m

Distance between proton
1 and electron, *d*_{2} = 1 ×10^{−10}
m

Distance between proton
2 and electron, *d*_{3} = 1 × 10^{−10}
m

The potential energy at infinity is zero.

Potential energy of the system,

Therefore, the potential energy of the system is −19.2 eV.

#### Page No 89:

#### Question 2.20:

Two charged conducting spheres of radii *a* and *b* are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

#### Answer:

Let *a* be the radius of a sphere A, *Q*_{A}_{ }be the charge on the sphere, and *C*_{A} be the capacitance of the sphere. Let *b* be the radius of a sphere B, *Q*_{B} be the charge on the sphere, and *C*_{B} be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (*V*) will become equal.

Let *E*_{A}be the electric field of sphere A and *E*_{B} be the electric field of sphere B. Therefore, their ratio,

Putting the value of (2) in (1), we obtain

Therefore, the ratio of electric fields at the surface is.

A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.

#### Page No 89:

#### Question 2.21:

Two charges *−q *and *+q *are located at points (0, 0, − *a*) and (0, 0, *a*), respectively.

**(a)** What is the electrostatic potential at the points?

**(b)** Obtain the dependence of potential on the distance *r *of a point from the origin when *r*/*a *>> 1.

**(c)** How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the *x*-axis? Does the answer change if the path of the test charge between the same points is not along the *x*-axis?

#### Answer:

**(a) **Zero at both the points

Charge − *q* is located at (0, 0, − *a*) and charge + *q* is located at (0, 0, *a*). Hence, they form a dipole. Point (0, 0, *z*) is on the axis of this dipole and point (*x*, *y*, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (*x*, *y*, 0) is zero. Electrostatic potential at point (0, 0, *z*) is given by,

Where,

= Permittivity of free space

*p* = Dipole moment of the system of two charges = 2*qa*

**(b)** Distance *r* is much greater than half of the distance between the two charges. Hence, the potential (*V*) at a distance *r* is inversely proportional to square of the distance i.e.,

**(c) **Zero

The answer does not change if the path of the test is not along the *x*-axis.

A test charge is moved from point (5, 0, 0) to point (−7, 0, 0) along the *x*-axis. Electrostatic potential (*V*_{1}) at point (5, 0, 0) is given by,

${V}_{1}=\frac{-q}{4\mathrm{\pi}{\in}_{0}}\frac{1}{\sqrt{(5-0{)}^{2}}+(-a{)}^{2}}+\frac{q}{4\mathrm{\pi}{\in}_{0}}\frac{1}{\sqrt{(5-0{)}^{2}+}(a{)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-q}{4\mathrm{\pi}{\in}_{0}\sqrt{25+{\mathrm{a}}^{2}}}+\frac{q}{4\mathrm{\pi}{\in}_{0}\sqrt{25+{\mathrm{a}}^{2}}}\phantom{\rule{0ex}{0ex}}=0$

Electrostatic potential, *V*_{2}, at point (− 7, 0, 0) is given by,

Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the *x*-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

#### Page No 89:

#### Question 2.22:

Figure
2.34 shows a charge array known as an *electric
quadrupole*. For a point on the axis
of the quadrupole, obtain the dependence of potential on *r
*for *r*/*a
*>> 1, and contrast your
results with that due to an electric dipole, and an electric monopole
(i.e., a single charge).

#### Answer:

Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.

A point is located at
P, which is *r* distance away from point Y.

The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges.

Charge +*q *placed
at point X

Charge −2*q*
placed at point Y

Charge +*q* placed
at point Z

XY = YZ = *a*

YP =* r*

PX = *r* + *a*

PZ = *r* − *a*

Electrostatic potential caused by the system of three charges at point P is given by,

Since,

is taken as negligible.

It can be inferred that potential,

However, it is known that for a dipole,

And, for a monopole,

#### Page No 89:

#### Question 2.23:

An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

#### Answer:

Total required
capacitance, *C* = 2 µF

Potential
difference, *V*
= 1 kV = 1000 V

Capacitance
of each capacitor, *C*_{1}
= 1µF

Each
capacitor can withstand a potential difference, *V*_{1}
= 400 V

Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as

Hence, there are three capacitors in each row.

Capacitance of each row

Let
there are *n*
rows, each having three capacitors, which are connected in parallel.
Hence, equivalent capacitance of the circuit is given as

Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required for the given arrangement.

#### Page No 89:

#### Question 2.24:

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

#### Answer:

Capacitance of a
parallel capacitor, *V* = 2 F

Distance between the
two plates, *d* = 0.5 cm = 0.5 ×
10^{−2} m

Capacitance of a parallel plate capacitor is given by the relation,

Where,

= Permittivity of free space = 8.85 ×
10^{−12} C^{2} N^{−1} m^{−2}

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.

#### Page No 90:

#### Question 2.25:

Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.

#### Answer:

Capacitance of
capacitor *C*_{1 }is 100 pF.

Capacitance of
capacitor *C*_{2 }is 200 pF.

Capacitance of
capacitor *C*_{3 }is 200 pF.

Capacitance of
capacitor *C*_{4 }is 100 pF.

Supply potential, *V*
= 300 V

Capacitors *C*_{2}
and *C*_{3} are connected in series. Let their
equivalent capacitance be

Capacitors *C*_{1}
and *C’ *are in parallel. Let their equivalent capacitance
be

are
connected in series. Let their equivalent capacitance be *C*.

Hence, the equivalent capacitance of the circuit is

Potential difference across =

Potential difference
across *C*_{4} =* V*_{4}

Charge on

*Q*_{4}=
*CV*

Hence, potential
difference, *V*_{1}, across *C*_{1} is 100
V.

Charge on *C*_{1}
is given by,

*C*_{2}
and *C*_{3} having same capacitances have a potential
difference of 100 V together. Since *C*_{2} and *C*_{3}
are in series, the potential difference across *C*_{2 }and
*C*_{3 }is given by,

*V*_{2}*
= V*_{3}* = *50 V

Therefore, charge on *C*_{2}
is given by,

And charge on *C*_{3
}is given by,

Hence, the equivalent capacitance of the given circuit is

#### Page No 90:

#### Question 2.26:

The
plates of a parallel plate capacitor have an area of 90 cm^{2}
each and are separated by 2.5 mm. The capacitor is charged by
connecting it to a 400 V supply.

**(a)** How
much electrostatic energy is stored by the capacitor?

**(b)** View
this energy as stored in the electrostatic field between the plates,
and obtain the energy per unit volume *u*.
Hence arrive at a relation between *u
*and the magnitude of electric field
*E *between
the plates.

#### Answer:

Area of the plates of a parallel plate capacitor, *A* = 90 cm^{2
}= 90 × 10^{−4
}m^{2}

Distance between the plates, *d* = 2.5 mm = 2.5 ×
10^{−3 }m

Potential difference across the plates, *V* = 400 V

**(a) **Capacitance of the capacitor is given by the relation,

Electrostatic energy stored in the capacitor is given by the relation,

Where,

= Permittivity of free space = 8.85 ×
10^{−12} C^{2} N^{−1} m^{−2}

Hence, the electrostatic energy stored by the capacitor is

**(b) **Volume of the given capacitor,

Energy stored in the capacitor per unit volume is given by,

Where,

=
Electric intensity = *E*

#### Page No 90:

#### Question 2.27:

A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

#### Answer:

Capacitance of a charged capacitor,

Supply voltage, *V*_{1}
= 200 V

Electrostatic energy
stored in *C*_{1 }is given by,

Capacitance of an uncharged capacitor,

When *C*_{2}
is connected to the circuit, the potential acquired by it is *V*_{2}.

According to the
conservation of charge, initial charge on capacitor *C*_{1}
is equal to the final charge on capacitors, *C*_{1} and
*C*_{2}.

Electrostatic energy for the combination of two capacitors is given by,

Hence, amount of
electrostatic energy lost by capacitor *C*_{1}

= *E*_{1}
− *E*_{2}

= 0.08 − 0.0533 = 0.0267

= 2.67 ×
10^{−2} J

#### Page No 90:

#### Question 2.28:

Show
that the force on each plate of a parallel plate capacitor has a
magnitude equal to (½) *QE*,
where *Q *is
the charge on the capacitor, and *E *is
the magnitude of electric field between the plates. Explain the
origin of the factor ½.

#### Answer:

Let *F* be the
force applied to separate the plates of a parallel plate capacitor by
a distance of Δ*x*.
Hence, work done by the force to do so = *F**Δ**x*

As a result, the
potential energy of the capacitor increases by an amount given as
*uA**Δ**x*.

Where,

*u* = Energy
density

*A* = Area of each
plate

*d* = Distance
between the plates

*V* = Potential
difference across the plates

The work done will be equal to the increase in the potential energy i.e.,

Electric intensity is given by,

However, capacitance,

Charge on the capacitor is given by,

*Q* = *CV*

The physical origin of
the factor,
,
in the force formula lies in the fact that just outside the
conductor, field is *E* and inside it is zero. Hence, it is the
average value,
,
of the field that contributes to the force.

#### Page No 90:

#### Question 2.29:

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show

that the capacitance of a spherical capacitor is given by

where
*r*_{1}
and *r*_{2}
are the radii of outer and inner spheres, respectively.

#### Answer:

Radius of the outer
shell = *r*_{1}

Radius of the inner
shell = *r*_{2}

The inner surface of
the outer shell has charge +*Q*.

The outer surface of
the inner shell has induced charge −*Q*.

Potential difference between the two shells is given by,

Where,

= Permittivity of free space

Hence, proved.

#### Page No 91:

#### Question 2.30:

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

**(a)** Determine
the capacitance of the capacitor.

**(b)** What
is the potential of the inner sphere?

**(c)** Compare
the capacitance of this capacitor with that of an isolated sphere of
radius 12 cm. Explain why the latter is much smaller.

#### Answer:

Radius of the inner sphere, = 12 cm = 0.12 m

Radius of the outer sphere, = 13 cm = 0.13 m

Charge on the inner sphere,

Dielectric constant of a liquid,

**(a) **

Where,

= Permittivity of free space =

Hence, the capacitance of the capacitor is approximately .

**(b) **Potential of
the inner sphere is given by,

Hence, the potential of the inner sphere is .

**(c) **Radius of an
isolated sphere, *r *= 12 ×
10^{−2} m

Capacitance of the sphere is given by the relation,

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.

#### Page No 91:

#### Question 2.31:

Answer carefully:

**(a) ** Two
large conducting spheres carrying charges *Q*_{1}
and *Q*_{2}
are brought close to each other. Is the magnitude of electrostatic
force between them exactly given by *Q*_{1}*
Q*_{2}/4π*r
*^{2},
where *r *is
the distance between their centres?

**(b)** If
Coulomb’s law involved 1/*r*^{3}
dependence (instead of 1/*r*^{2}),
would Gauss’s law be still true?

**(c)** A small
test charge is released at rest at a point in an electrostatic field
configuration. Will it travel along the field line passing through
that point?

**(d)** What is
the work done by the field of a nucleus in a complete circular orbit
of the electron? What if the orbit is elliptical?

**(e)** We know
that electric field is discontinuous across the surface of a charged
conductor. Is electric potential also discontinuous there?

**(f)** What
meaning would you give to the capacitance of a single conductor?

**(g)** Guess a
possible reason why water has a much greater dielectric constant (=
80) than say, mica (= 6).

#### Answer:

**(a) **The force between two conducting spheres is not exactly
given by the expression, *Q*_{1}*
Q*_{2}/4π*r
*^{2},
because there is a non-uniform charge distribution on the spheres.

**(b) **Gauss’s law will not be true,
if Coulomb’s law involved 1/*r*^{3
}dependence, instead of1/*r*^{2},
on *r*.

**(c) **Yes,

If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.

**(d) **Whenever the electron completes an
orbit, either circular or elliptical, the work done by the field of a
nucleus is zero.

**(e) **No

Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

**(f) **The capacitance of a single
conductor is considered as a parallel plate capacitor with one of its
two plates at infinity.

**(g) **Water has an unsymmetrical space as
compared to mica. Since it has a permanent dipole moment, it has a
greater dielectric constant than mica.

#### Page No 91:

#### Question 2.32:

A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

#### Answer:

Length of a co-axial
cylinder,* l* = 15 cm = 0.15 m

Radius of outer
cylinder, *r*_{1} = 1.5 cm = 0.015 m

Radius of inner
cylinder, *r*_{2} = 1.4 cm = 0.014 m

Charge on the inner
cylinder, *q* = 3.5 µC = 3.5 ×
10^{−6}
C

Where,

= Permittivity of free space =

Potential difference of the inner cylinder is given by,

#### Page No 91:

#### Question 2.33:

A
parallel plate capacitor is to be designed with a voltage rating 1
kV, using a material of dielectric constant 3 and dielectric strength
about 10^{7}
Vm^{−1}.
(Dielectric strength is the maximum electric field a material can
tolerate without breakdown, i.e., without starting to conduct
electricity through partial ionisation.) For safety, we should like
the field never to exceed, say 10% of the dielectric strength. What
minimum area of the plates is required to have a capacitance of 50
pF?

#### Answer:

Potential rating of a
parallel plate capacitor, *V* = 1 kV = 1000 V

Dielectric constant of a material, =3

Dielectric strength =
10^{7} V/m

For safety, the field intensity never exceeds 10% of the dielectric strength.

Hence, electric field
intensity, *E* = 10% of 10^{7 }= 10^{6} V/m

Capacitance of the
parallel plate capacitor, *C* = 50 pF = 50 ×
10^{−12} F

Distance between the plates is given by,

Where,

*A* = Area of each
plate

= Permittivity of free space =

Hence, the area of each
plate is about 19 cm^{2}.

#### Page No 91:

#### Question 2.34:

Describe schematically the equipotential surfaces corresponding to

**(a)** a
constant electric field in the *z*-direction,

**(b)** a field
that uniformly increases in magnitude but remains in a constant (say,
*z*)
direction,

**(c)** a
single positive charge at the origin, and

**(d)** a
uniform grid consisting of long equally spaced parallel charged wires
in a plane.

#### Answer:

**(a) **Equidistant
planes parallel to the *x*-*y* plane are the equipotential
surfaces.

**(b) **Planes parallel to the *x-y *plane are the
equipotential surfaces with the exception that when the planes get
closer, the field increases.

**(c) **Concentric
spheres centered at the origin are equipotential surfaces.

**(d) **A periodically varying shape near the given grid is the
equipotential surface. This shape gradually reaches the shape of
planes parallel to the grid at a larger distance.

#### Page No 92:

#### Question 2.35:

In
a Van de Graaff type generator a spherical metal shell is to be a 15
× 10^{6}
V electrode. The dielectric strength of the gas surrounding the
electrode is 5 × 10^{7}
Vm^{−1}.
What is the minimum radius of the spherical shell required? (You will
learn from this exercise why one cannot build an electrostatic
generator using a very small shell which requires a small charge to
acquire a high potential.)

#### Answer:

Potential difference, *V*
= 15 × 10^{6} V

Dielectric strength of
the surrounding gas = 5 ×
10^{7} V/m

Electric field
intensity, *E* = Dielectric strength = 5 ×
10^{7} V/m

Minimum radius of the spherical shell required for the purpose is given by,

Hence, the minimum radius of the spherical shell required is 30 cm.

#### Page No 92:

#### Question 2.36:

A
small sphere of radius *r*_{1}
and charge *q*_{1}
is enclosed by a spherical shell of radius *r*_{2}
and charge *q*_{2}.
Show that if *q*_{1}
is positive, charge will necessarily flow from the sphere to the
shell (when the two are connected by a wire) no matter what the
charge *q*_{2}
on the shell is.

#### Answer:

According to Gauss’s
law, the electric field between a sphere and a shell is determined by
the charge *q*_{1}
on a small sphere. Hence, the potential difference, *V*,
between the sphere and the shell is independent of charge *q*_{2}._{
}For positive charge *q*_{1},
potential difference *V *is
always positive.

#### Page No 92:

#### Question 2.37:

Answer the following:

**(a)** The top
of the atmosphere is at about 400 kV with respect to the surface of
the earth, corresponding to an electric field that decreases with
altitude. Near the surface of the earth, the field is about 100 Vm^{−1}.
Why then do we not get an electric shock as we step out of our house
into the open? (Assume the house to be a steel cage so there is no
field inside!)

**(b)** A man
fixes outside his house one evening a two metre high insulating slab
carrying on its top a large aluminium sheet of area 1m^{2}.
Will he get an electric shock if he touches the metal sheet next
morning?

**(c)** The
discharging current in the atmosphere due to the small conductivity
of air is known to be 1800 A on an average over the globe. Why then
does the atmosphere not discharge itself completely in due course and
become electrically neutral? In other words, what keeps the
atmosphere charged?

**(d)** What
are the forms of energy into which the electrical energy of the
atmosphere is dissipated during a lightning? (Hint: The earth has an
electric field of about 100 Vm^{−1}
at its surface in the downward direction, corresponding to a surface
charge density = −10^{−9}
C m^{−2}.
Due to the slight conductivity of the atmosphere up to about 50 km
(beyond which it is good conductor), about + 1800 C is pumped every
second into the earth as a whole. The earth, however, does not get
discharged since thunderstorms and lightning occurring continually
all over the globe pump an equal amount of negative charge on the
earth.)

#### Answer:

**(a) **We do not get an electric shock as we step out of our
house because the original equipotential surfaces of open air
changes, keeping our body and the ground at the same potential.

**(b) **Yes, the man will get an electric shock if he touches the
metal slab next morning. The steady discharging current in the
atmosphere charges up the aluminium sheet. As a result, its voltage
rises gradually. The raise in the voltage depends on the capacitance
of the capacitor formed by the aluminium slab and the ground.

**(c) **The occurrence of thunderstorms and lightning charges the
atmosphere continuously. Hence, even with the presence of discharging
current of 1800 A, the atmosphere is not discharged completely. The
two opposing currents are in equilibrium and the atmosphere remains
electrically neutral.

**(d) **During lightning and thunderstorm, light energy, heat
energy, and sound energy are dissipated in the atmosphere.

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