Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 2 - Electrostatic Potential And Capacitance

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Page No 10:

Question 1:

A capacitor of 4 μ F is connected as shown in the circuit in the given figure. The internal resistance of the battery is 0.5 â„¦ . The amount of charge on the capacitor plates will be

(a) 0
(b) 4 μ C
(c) 16 μ C
(d) 8 μ C

Answer:

Capacitance of the capacitor is given by,
C = 4 μF
The charge stored on the capacitor is given by the formula,
Q = CV
Now to calculate the potential difference across the capacitor.
The total current in the circuit is given by,
$I = \frac{V}{R + r} = \frac{2.5}{2 + 0.5} = 1 A$
The potential difference across the 2 Î resistor is 2 V.
The potential difference across the capacitor is also 2V.
So the charge stored within the capacitor is given by,
$Q = C V = 4 \times 10^{- 6} \times 2 = 8 \times 10^{- 6} C$
The charge across the capacitor is 8 μC.

Hence, the correct answer is option (d).

Page No 10:

Question 2:

A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge

(a) remains a constant because the electric field is uniform.

(b) increases because the charge moves along the electric field.

(d) decreases because the charge moves opposite to the electric field.

Answer:

When a positively charged particle is released from rest in a uniform field, it moves along the electric field, i.e., from higher potential to lower potential. That is electric potential energy of the positively charged particle decreases when it moves along the electric field.
Hence, the correct answer is option (c).

Page No 11:

Question 3:

In the given figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

(a) The work done in Fig. (i) is the greatest.

(b) The work done in Fig. (ii) is least.

(d) The work done in Fig. (iii) is greater than Fig. (ii)but equal to that in Fig. (i).

Answer:

The work done in moving a charged particle is given by,
W = q ΔV
where q is the charge on the particle and ΔV is the potential difference.
We can see that ΔV = 20 V is the same in all three figures, so the work done is also equal in all three configurations.

Hence, the correct answer is option (c).

Page No 11:

Question 4:

The electrostatic potential on the surface of a charged conducting sphere is 100V. Two statements are made in this regard:
S₁ : At any point inside the sphere, electric intensity is zero.
S₂ : At any point inside the sphere, the electrostatic potential is 100V.

Which of the following is a correct statement?

(a) S₁ is true but S₂is false.

(b) Both S₁ & S₂ are false.

(d) S₁ is true, S₂ is also true but the statements are independent.

Answer:

The electric field inside the sphere is zero.
The electric potential V is given by the formula,
$E = - \frac{d V}{d r} \Rightarrow 0 = - \frac{d V}{d r} \Rightarrow V is constant inside the sphere .$

Hence, the correct answer is option (c).

Page No 11:

Question 5:

Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately

(a) spheres.

(b) planes.

(d) ellipsoids.

Answer:

The collection of these charges can be considered as a point charge with respect to a very far point. So the equipotential surface due to such a point charge is spherical in shape.
â€‹Hence, the correct answer is option (a).

Page No 12:

Question 6:

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d₁ and dielectric constant k₁ and the other has thickness d₂ and dielectric constant k₂as shown in the given figure. This arrangement can be thought as a dielectric slab of thickness d (= d₁+ d₂) and effective dielectric constant k. The k is

$(a) \frac{k_{1} d_{1} + k_{2} d_{2}}{d_{1} + d_{2}} (b) \frac{k_{1} d_{1} + k_{2} d_{2}}{k_{1} + k_{2}} (c) \frac{k_{1} k_{2} (d_{1} + d_{2})}{(k_{1} d_{1} + k_{2} d_{2})} (d) \frac{2 k_{1} k_{2}}{k_{1} + k_{2}}$

Answer:

The two capacitors can be thought of as they are connected in series.
The net capacitance is given by C_net
$\frac{1}{C_{n e t}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$
where C₁and C₂are given by the expressions
$C_{1} = \frac{k_{1} ε_{o} A}{d_{1}} & C_{2} = \frac{k_{2} ε_{o} A}{d_{2}}$
$C_{n e t} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} C_{n e t} = \frac{k_{1} k_{2} ε_{o} A}{k_{1} d_{2} + k_{2} d_{1}}$
We can compare the standard value of the capacitance with the above-calculated value of the net capacitance and the effective value of the dielectric constant can be found accordingly.
$k_{n e t} = \frac{k_{1} k_{2} (d_{1} + d_{2})}{k_{1} d_{2} + k_{2} d_{1}}$
Hence, the correct answer is option (c).

Page No 12:

Question 7:

Consider a uniform electric field in the

\hat{z}

direction. The potential is a constant

(a) in all space.

(b) for any x for a given z.

(d) on the x-y plane for a given z.

Answer:

Since, electric field along z-direction is constant. i.e.,
$E_{Z} = constant$
But in x and y-direction, there is no electric field, i.e.,
$E_{x} = 0 and E_{y} = 0$
But, electric field is defined as:
$E = - \frac{d V}{d r}$
So, potential along x and y direction will be constant but in z-direction it will vary.

Hence, the correct options are (b), (c), and (d).

Page No 12:

Question 8:

Equipotential surfaces

(a) are closer in regions of large electric fields compared to regions of lower electric fields.

(b) will be more crowded near sharp edges of a conductor.

(d) will always be equally spaced.

Answer:

The electric field intensity is defined as the negative gradient of the electric potential.

$E = - \frac{d V}{d r}$

It means, the electric field intensity E is inversely proportional to the separation between equipotential surfaces.
So, equipotential surfaces are closer in regions of large electric fields.

Hence, the correct options are (a), (b) and (c).

Page No 12:

Question 9:

The work done to move a charge along an equipotential from A to B

(a) cannot be defined as

- \int_{A}^{B} E . d l

(b) must be defined as

- \int_{A}^{B} E . d l

(d) can have a non-zero value.

Answer:

Work done in moving a charged particle from point A to the point B is given by the formula:
$W = q ∆ V$
where ΔV is the potential difference between the two points.
In an equipotential surface, the potential is constant throughout, So the Work done will be zero.
The value of the potential difference is given by the formula $∆ V = - \int_{A}^{B} E . d l$

Hence, the correct options are (b) and (c).

Page No 12:

Question 10:

â€‹In a region of constant potential

(a) the electric field is uniform

(b) the electric field is zero

(d) the electric field shall necessarily change if a charge is placed outside the region.

Answer:

A region of constant potential is known as the equipotential surface and electric field E can be expressed by the formula $E = - \frac{d V}{d r}$
If V is a constant then the value of the electric field will be zero.

So, if E is zero then there must be a net-zero charge inside that region.

Hence, the correct options are (b) and (c).

Page No 12:

Question 11:

In the circuit shown in the given figure. initially key K₁ is closed and key K₂ is open. Then K₁ is opened and K₂ is closed (order is important). [Take Q₁′ and Q₂′ as charges on C₁ and C₂ and V₁and V₂as voltage respectively.]

Then
(a) charge on C₁ gets redistributed such that V₁= V₂
(b) charge on C₁gets redistributed such that Q₁′ = Q₂′
(c) charge on C₁ gets redistributed such that C₁V₁ + C₂V₂ = C₁E
(d) charge on C₁ gets redistributed such that Q₁′ + Q₂′ = Q

Answer:

Initially, the capacitor C₁gets charged when key K₁is closed and key K₂is open. Now at this point capacitor, C₁is charged and C₂is uncharged.
When key K₁is opened and K₂is closed the charge stored in the capacitor C₁gets redistributed between C₁and C₂till their potential becomes equal V₁= V₂
The charge also gets redistributed such that Q₁′ + Q₂′ = Q
Hence, the correct options are (a) and (d).

Page No 13:

Question 12:

â€‹If a conductor has a potential V ≠ 0 and there are no charges anywhere else outside, then

(a) there must be charges on the surface or inside itself.

(b) there cannot be any charge in the body of the conductor.

(d) there must be charges inside the surface.

Answer:

The potential on a body is due to the charge present on the surface or inside the conductor.
Electric field inside a conductor is zero. So, according to Gauss law ,
$\oint E . d s = \frac{q_{i n}}{ε_{o}} \Rightarrow q_{i n} = 0$
Hence, there can not be any charge in the body of a conductor.

Hence, the correct answer is option (a) and option (b).

Page No 13:

Question 13:

A parallel plate capacitor is connected to a battery as shown in the given Figure. Consider two situations:

A: Key K is kept closed and plates of capacitors are moved apart using insulating handle.

B: Key K is opened and plates of capacitors are moved apart using insulating handle.

â€‹Choose the correct option(s).

(a) In A : Q remains same but C changes.

(b) In B : V remains same but C changes.

(d) In B : Q remains same and hence V changes.

Answer:

As the battery remains connected in part A, So the potential V remains the same only Q changes.

In B the battery is removed So the value of the charge on each capacitor remains the same but V changes.

Hence, the correct answers are option (c) and (d).

Page No 13:

Question 14:

â€‹Consider two conducting spheres of radii R₁and R₂ with R₁> R₂. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Answer:

Charge density is given by the charge per unit area, mathematically it is expressed as $σ = \frac{Q}{4 π R^{2}}$

where R is the radius of the sphere Q is the charge on the sphere.

It can be seen from the formula that Charge density is inversely proportional to the square of the radius of the sphere.

So, it can be concluded from the discussion that the smaller sphere has a large surface charge density

Page No 14:

Question 15:

â€‹Do free electrons travel to region of higher potential or lower potential?

Answer:

Electrons are negatively charged particles and they experience an electrostatic force in a direction opposite to the direction of the electric field.

We know that the Electric potential decreases along the direction of the Electric field, So it can be concluded that electrons move from a lower potential region to a higher potential region

Page No 14:

Question 16:

Can there be a potential difference between two adjacent conductors carrying the same charge? â€‹

Answer:

Yes, there can be a potential difference between the two adjacent conductors carrying the same charge, if their sizes are different.

$V = \frac{Q}{C}$ , the potential difference is inversely proportional to the capacitance of a conductor.

Capacitance does depend on the size of the conductor and it varies with the shape of the conductor.

Page No 14:

Question 17:

â€‹Can the potential function have a maximum or minimum in free space?

Answer:

The potential function does not have a maximum or minimum in free space because in the absence of free space the phenomenon of electric field or potential leakage cannot be prevented.

Page No 14:

Question 18:

A test charge q is made to move in the electric field of a point charge Q along two different closed paths in the given figure. First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?

Answer:

The electrostatic field is conservative in nature and the work done in moving a charge particle is independent of the path followed by the particle, It rather depends on the initial and final position of the charged particle.
So the work done in moving the charged particle in a closed loop is zero, Thus it can be concluded that the work done in both the cases is zero.

Page No 14:

Question 19:

â€‹Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

Answer:

Let's consider a closed spherical surface with no charge and a potential gradient when we move from the center towards the surface. Let's assume it to be an equipotential surface. Due to this potential gradient, there is a field which exists in this region, and the magnitude of the field is given by the formula

$E = - \frac{d V}{d r}$

The field must be originating due to some electric charge which is present inside the spherical surface, This result is in contradiction to the original assumption. Thus it can be concluded that the entire volume is equipotential.

Page No 14:

Question 20:

â€‹A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

Answer:

The capacitance of the capacitor is given by the expression $C = \frac{k ε_{o} A}{d}$

where k is the dielectric constant of the dielectric material inserted between the plates of the parallel plate capacitor.

When the battery is disconnected the charge on the plates of the parallel plate capacitor remains the same.

The dielectric slab is also removed so the capacitance of the capacitor decreases.

Energy stored in the parallel plate capacitor is given by the formula $U = \frac{q^{2}}{2 C}$ and if the capacitance decreases the Energy stored within the plates of the capacitor increases.

Potential difference V is given by $V = \frac{q}{C}$ , As the charge q is constant and C decreases this will increase the value of the Potential differnce across the plates of the parallel plate capacitor.

The electric field between the plate of the parallel plate capacitor is given by $E = \frac{V}{d}$ , It can be concluded that if Potential differnce V increases the value of the Electric field E also increases.

Page No 14:

Question 21:

â€‹Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Answer:

The relationship between the Electric field E and the electric potential V is given by the expression $E = - \frac{d V}{d r}$
The expression physically means that the electric potential decreases along the direction of the electric field. If we have a positively charged conductor its field would be pointing in a direction away from the charged conductor and potential due to this charged conductor will decrease in the direction of the field and will vanish at infinity. So any point in between the conducting sphere and infinity will be having a non zero potential intermediate that of the charged body and infinity.

Page No 14:

Question 22:

â€‹Calculate potential energy of a point charge –q placed along the axis due to a charge +Q uniformly distributed along a ring of radius R. Sketch P.E. as a function of axial distance z from the centre of the ring. Looking at graph, can you see what would happen if –q is displaced slightly from the centre of the ring (along the axis)?

Answer:

Electric potential is given by the formula $V = \frac{1}{4 π ε_{o}} \frac{Q}{\sqrt{z^{2} + R^{2}}}$

Here the axis of the ring is along the z-direction.

The potential energy of the charge -q is given by U = $\frac{- q}{4 π ε_{o}} \frac{Q}{\sqrt{z^{2} + R^{2}}}$

The variation in potential energy as a function of the distance z along the axis of the ring is given by the above expression.

The graph is below:
Here, in the image (R = a)

for z = 0, U = $\frac{- q Q}{4 π ε_{o} R}$

for $z \to \infty, U = 0$

Page No 14:

Question 23:

â€‹Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.

Answer:

Consider the ring as shown in the figure below with radius R and charge Q distributed uniformly over it.

Consider a small charge element dq at a point P on the circumference of the ring.

Let's calculate the value of the electric potential at a point P on the axis of the ring which at a distance z from the center of the ring.

$d V = \frac{d q}{4 π ε_{o} \sqrt{R^{2} + z^{2}}}$ is the value of the electric potential at a point P due to a charge dq.

In order to calculate the potential due to the complete ring, we will carry out the integration within the limits

$\int_{}^{} d V = \int_{}^{} \frac{d q}{4 π ε_{o} \sqrt{R^{2} + z^{2}}} V = \frac{1}{4 π ε_{o} \sqrt{R^{2} + z^{2}}} \int_{}^{} d q V = \frac{Q}{4 π ε_{o} \sqrt{R^{2} + z^{2}}}$

Page No 14:

Question 24:

â€‹Find the equation of the equipotentials for an infinite cylinder of radius r₀, carrying charge of linear density λ.

Answer:

$\int E_{r} . d s = \frac{λ l}{ε_{o}} \Rightarrow E_{r} 2 π r l = \frac{λ l}{ε_{o}} \Rightarrow E_{r} = \frac{λ}{2 π ε_{o} r}$
So, if r_o is the radius of the cylinder, then:
$V (r) - V (r_{o}) = - \int_{r_{o}}^{r} E_{r} d r = - \int_{r_{o}}^{r} \frac{λ}{2 π ε_{o} r} d r = = - \frac{λ}{2 π ε_{o}} \int_{r_{o}}^{r} \frac{1}{r} \cdot d r = - \frac{λ}{2 π ε_{o}} \ln \frac{r}{r_{o}}$

$\Rightarrow \ln \frac{r}{r_{o}} = \frac{- 2 π ε_{o}}{λ} [V (r) - V (r_{o})] \Rightarrow \frac{r}{r_{o}} = e^{\frac{- 2 π ε_{o} [V_{(r)} - V_{(r_{o})}]}{λ}} (using the property of \log) \Rightarrow r = r_{o} e^{\frac{- 2 π ε_{o} [V_{(r)} - V_{(r_{o})}]}{λ}}$

Page No 15:

Question 25:

â€‹Two point charges of magnitude +q and –q are placed at (–d/2, 0,0) and (d/2, 0,0), respectively. Find the equation of the equipoential surface where the potential is zero.

Answer:

Consider the required plane lies at a distance 'x' from the origin as shown in the figure.

The potential at the point P due to charges is given by, $\frac{1}{4 π ε_{o}} \frac{q}{\sqrt{{(x + \frac{d}{2})}^{2} + h^{2}}} - \frac{1}{4 π ε_{o}} \cdot \frac{q}{\sqrt{{(x - \frac{d}{2})}^{2} + h^{2}}}$
Net electric potential is zero, then $\frac{1}{\sqrt{{(x + \frac{d}{2})}^{2} + h^{2}}} = \frac{1}{\sqrt{{(x - \frac{d}{2})}^{2} + h^{2}}}$
On solving we get, x = 0 i.e., y – z plane.

Page No 15:

Question 26:

â€‹A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where α = 2V^–1 .A similar capacitor with no dielectric is charged to U_o = 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Answer:

On connecting the two given capacitors, let the final voltage be U.
If the capacity of capacitor without dielectric is C,
then charge on this capacitor is q₁ = CU
The other capacitor with dielectric has capacity εC.
Therefore, charge on it is q₂ = εCU
As ε = αU, therefore q₂ = αCU²
The initial charge on the capacitor (without dielectric) that was charged q_o = CU_o
From the conservation of charge, q_o = q₁+ q₂
or CU_o = CU + αCU²
$\Rightarrow α U^{2} + U - U_{o} = 0 \Rightarrow U = \frac{- 1 \pm \sqrt{1 + 4 α U_{o}}}{2 α} (Roots of a quadratic equation)$
On solving for U_o= 78 V and $\propto = 2 V^{- 1}$ , we get:
$\Rightarrow U = \frac{- 1 \pm \sqrt{1 + (4 \times 2 \times 78)}}{2 \times 2} = \frac{- 1 \pm \sqrt{625}}{4}$
U can never be negative, so taking only positive value of U,
$U = \frac{\sqrt{625} - 1}{4} = \frac{24}{4} = 6 V$ .

Page No 15:

Question 27:

â€‹A capacitor is made of two circular plates of radius R each, separated by a distance d<<R. The capacitor is connected to a constant voltage. A thin conducting disc of radius r<<R and thickness t<<r is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is m.

Answer:

As shown in the diagram the radius of the circular capacitor is R. Distance between the plates is d. Potential difference between the plates is v. So, electric field between the plates $E = \frac{v}{d}$
Due to which charge is transferred to the disc.
Let charge on the disc (of radius r and thickness t) placed on the lower plate be q', then
$q' = ε_{0} \frac{v}{d} \times π r^{2} (∵ q = c v, and c = \frac{ε_{0} A'}{d} A' = π r^{2})$
Now force acting on the disc is due to this charge and the electric field,
$\begin{array}{rcl} q' \times E & = & ε_{0} \frac{v}{d^{2}} \times π r^{2} \times \frac{v}{d} \\ = & ε_{0} \frac{v^{2}}{d^{2}} \times π r^{2} \end{array}$
If the disc to be lifted,
$q' \times E = m g \Rightarrow ε_{0} \frac{v^{2}}{d^{2}} \times π r^{2} = m g \Rightarrow v = \sqrt{\frac{m g d^{2}}{ε_{0} π r^{2}}}$
This is the required voltage to lift the disc up.

Page No 15:

Question 28:

â€‹(a) In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3) e] and two down quarks [charges –(1/3) e]. Assume that they have a triangle configuration with side length of the order of 10^–15m. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.

(b) Repeat above exercise for a proton which is made of two up and one down quark.

Answer:

(a) Up quarks and down quarks are at three corners of the triangle as shown in the diagram:

So, the potential energy of the system is,
$\begin{array}{rcl} u & = & \frac{1}{4 π ε_{0}} \times \frac{\frac{2}{3} e \times (- \frac{1}{3} e)}{r} + \frac{1}{4 {πε}_{0}} \times \frac{\frac{2}{3} e \times (- \frac{1}{3} e)}{r} + \frac{1}{4 π ε_{0}} \times \frac{(- \frac{1}{3} e) \times (- \frac{1}{3} e)}{r} \\ = & \frac{e^{2}}{4 π ε_{0} r} [\frac{1}{9} - \frac{2}{9} - \frac{2}{9}] \\ = & - \frac{9 \times 10^{9} \times {(1.6 \times 10^{- 19})}^{2}}{10^{- 15}} \times \frac{1}{3} \\ = & - 7.68 \times 10^{- 14} J \\ = & - \frac{7.68 \times 10^{- 14}}{1.6 \times 10^{- 19}} e V \\ |u| & \approx & 0.48 MeV = 5.11 \times 10^{- 4} (m_{n} c^{2}) \end{array}$

(b) Similarly for a proton which is made of two up and one down quark:
There are two up quarks having charges, (2/3) e and one down quark having charge –(1/3) e, which are situated at the three corners of the equilateral triangle of side length, r = 10⁻¹⁵ m.

So, the potential energy of the system is,
$\begin{array}{rcl} u' & = & \frac{1}{4 π ε_{0}} \times \frac{\frac{2}{3} e \times (\frac{2}{3} e)}{r} + \frac{1}{4 {πε}_{0}} \times \frac{\frac{2}{3} e \times (- \frac{1}{3} e)}{r} + \frac{1}{4 π ε_{0}} \times \frac{(- \frac{1}{3} e) \times (\frac{2}{3} e)}{r} \\ = & \frac{e^{2}}{4 π ε_{0} r} [\frac{4}{9} - \frac{2}{9} - \frac{2}{9}] \\ = & 0 J = 0 MeV \end{array}$

Page No 15:

Question 29:

Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density $σ$ . They are brought in contact and separated. What will be new surface charge densities on them?

Answer:

When the spheres are not in contact:
Given surface charge density of both the spheres is σ.

So, Q₁ = σ × 4πR² .....(1)
and, Q₂= σ × 4π(2R)² = 4 × (σ × 4πR²)
= 4Q₁ .....(2)
Now, where the spheres are brought in contact, then charge gets exchanged and new charge on the spheres are $Q_{1}^{'} and Q_{2}^{'} .$
But the charge must be conserved,
So, $Q_{1}^{'} + Q_{2}^{'} = Q_{1} + Q_{2}$
$Q_{1}^{'} + Q_{2}^{'} = Q_{1} + Q_{2} \Rightarrow Q_{1}^{'} + Q_{2}^{'} = Q_{1} + 4 Q_{1} = 5 Q_{1} \Rightarrow Q_{1}^{'} + Q_{2}^{'} = 5 \times (σ \times 4 π R^{2}) . . . . . (3)$
After the contact, both the spheres must be at same potential,
So, $\frac{1}{4 π ε_{0}} \times \frac{Q_{1}^{'}}{R} = \frac{1}{4 π ε_{0}} \times \frac{Q_{2}^{'}}{2 R}$
$\Rightarrow Q_{2}^{'} = 2 Q_{1}^{'} \Rightarrow Q_{1}^{'} + Q_{2}^{'} = 3 Q_{1}^{'} = 5 \times (σ \times 4 π R^{2}) [From equation (3)] \Rightarrow Q_{1}^{'} = \frac{5}{3} \times (σ \times 4 π R^{2})$
$and Q_{2}^{'} = 2 Q_{1}^{'} = \frac{10}{3} \times (σ \times 4 π R^{2})$
So, charge density of smaller sphere,
$σ_{1}^{'} = \frac{\frac{5}{3} \times (σ \times 4 π R^{2})}{4 π R^{2}} = \frac{5}{3} σ$
Thus, charge density of large sphere,
$\begin{array}{rcl} σ_{2}^{'} & = & \frac{\frac{10}{3} \times (σ \times 4 π R^{2})}{4 π (2 R^{2})} = \frac{10}{3} \times \frac{1}{4} σ \\ = & \frac{5}{6} σ \end{array}$

Page No 15:

Question 30:

In the circuit shown in the given figure, initially K₁ is closed and K₂ is open. What are the charges on each capacitors.

Then K₁ was opened and K₂ was closed (order is important), What will be the charge on each capacitor now? [C = 1 μF]

Answer:

When K₁ is cleared and K₂is open:

Given: C₁ = 6 C, C₂ = 3 C and C = 1 μF
Potential difference (V) across the capacitor of capacitance C, $V \propto \frac{1}{C}$
So, $\frac{V_{1}}{V_{2}} = \frac{C_{2}}{C_{1}} = \frac{3 C}{6 C} = \frac{1}{2} . . . . . (1)$
Also, V₁ + V₂ = E = 9 V .....(2)
Solving (1) and (2) we get,
⇒ V₁ = 3 V and V₂ = 6 V
Charge on first capacitor,
Q₁ = C₁ × V₁= (6 × 3) μC = 18 μC
Charge on second capacitor,
Q₂ = C₂ × V₂ = (3 × 6) μC = 18 μC
Charge on third capacitor,
Q₃ = 0 (Since, the K₂ is open)
Now, when K₁ is open and K₂ is closed:

No Source to supply charge,
So, the charge Q₂ will be re-distributed.
$Q_{2} = Q_{2}^{'} + Q_{3}^{'} = 18 μC$
Also, C₂V' = C₃V' = Q₂
$\Rightarrow V^{'} = \frac{Q_{2}}{C_{2} + C_{3}} = \frac{18 μC}{3 μF + 3 μF} = 3 V$
$So, Q_{2}^{'} = C_{2} \times V' = 3 \times 3 = 9 μC and Q_{3}^{'} = C_{3} \times V' = 3 \times 3 = 9 μC$

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Question 31:

â€‹Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface

Answer:

Charge on the disc shown is Q.
Radius of the disc is R
So, density of the charge on the disc, $σ = \frac{Q}{π R^{2}}$
Let us take a small element circular ring of radius dr and thickness(dr) as shown in the diagram.
Charge on the ring is,
dQ = σ × (2πr)dr
So, potential at point P on the axis of the disc due to this ring,
$\begin{array}{rcl} d U & = & \frac{1}{4 π ε_{0}} \times \frac{d q}{\sqrt{r^{2} + z^{2}}} (∵ P Q = \sqrt{r^{2} + z^{2}}) \\ = & \frac{1}{4 π ε_{0}} \times \frac{σ 2 π r d r}{\sqrt{r^{2} + z^{2}}} \end{array}$
Thus, the potential dt a point P on the axis of the disc due to the complete disc,
$\begin{array}{rcl} U & = & \int_{0}^{R} \frac{1}{4 π ε_{0}} \times \frac{σ 2 π r d r}{\sqrt{r^{2} + z^{2}}} \\ = & \frac{2 π σ}{4 π ε_{0}} (\sqrt{R^{2} + z^{2} - z}) \\ = & \frac{2 π \frac{Q}{π R^{2}}}{4 π ε_{0}} (\sqrt{R^{2} + z^{2} - z}) \\ = & \frac{2 Q}{4 π ε_{0} R^{2}} (\sqrt{R^{2} + z^{2} - z}) \end{array}$
This is the required expression for potential on the axis of the disc.

Page No 15:

Question 32:

â€‹Two charges q₁and q₂are placed at (0, 0, d) and (0, 0, –d) respectively. Find locus of points where the potential a zero.

Answer:

In the given diagram two charges q₁ and q₂ are placed at points (0, 0, d) and (0, 0, d) on z-axis.

Let co-ordinate of the point where potential is zero be P ≡ (x, y, z)
So, distance between the charge q₁ and the point P,
$x_{1} = \sqrt{{(x - 0)}^{2} + {(y - 0)}^{2} + {(z - d)}^{2}}$
Similarly, the distance between the charge q₂ and the point P,
$x_{2} = \sqrt{{(x - 0)}^{2} + {(y - 0)}^{2} + {(z + d)}^{2}}$
Now, if potential due to charge q₁ and q₂ at point P be zero, then:
$\frac{1}{4 π ε_{0}} \times \frac{q_{1}}{x_{1}} + \frac{1}{4 π ε_{0}} \times \frac{q_{2}}{x_{2}} = 0$
$\Rightarrow \frac{q_{1}}{\sqrt{x^{2} + y^{2} + {(z - d)}^{2}}} + \frac{q_{2}}{\sqrt{x^{2} + y^{2} + {(z + d)}^{2}}} = 0 \Rightarrow \frac{q_{1}}{\sqrt{x^{2} + y^{2} + {(z - d)}^{2}}} = - \frac{q_{2}}{\sqrt{x^{2} + y^{2} + {(z + d)}^{2}}}$
(It is clear that q₁ and q₂ must be of opposite sign.)
Now squaring both side we get,
$\frac{q_{1}^{2}}{x^{2} + y^{2} + {(z - d)}^{2}} = \frac{q_{2}^{2}}{x^{2} + y^{2} + {(z + d)}^{2}} \Rightarrow (x^{2} + y^{2} + z) + \frac{{(\frac{q_{1}}{q_{2}})}^{2} + 1}{{(\frac{q_{1}}{q_{2}})}^{2} - 1} (2 z d) + d^{2} = 0$
In co-ordinate system, the above equation represents the locus of a sphere of centre
$C \equiv (0, 0, - 2 d [\frac{q_{1}^{2} + q_{2}^{2}}{q_{1}^{2} - q_{2}^{2}}])$

Page No 15:

Question 33:

â€‹Two charges –q each are separated by distance 2d. A third charge +q is kept at mid point O. Find potential energy of +q as a function of small distance x from O due to – q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.

Answer:

As shown in the diagram, two charges each of charge –q placed at A and B and –q charge is placed at O.

Now, if the charge +q is displaced by a distance x as shown in diagram, then potential energy of the system be,
$\begin{array}{rcl} U & = & \frac{1}{4 π ε_{0}} [\frac{(- q) \times (+ q)}{(d - x)} + \frac{(- q) \times (+ q)}{(d + x)}] \\ = & - \frac{1}{4 π ε_{0}} \times \frac{q^{2} [(d + x) + (d - x)]}{(d - x) (d + x)} \\ = & - \frac{1}{4 π ε_{0}} \times \frac{q^{2} \times 2 d}{d^{2} - x^{2}} \end{array}$
Now, differentiating the potential energy with respect to x, we get:
$\begin{array}{rcl} \frac{d U}{d x} & = & \frac{d}{d x} [- \frac{1}{4 π ε_{0}} \times \frac{q^{2} \times 2 d}{d^{2} - x^{2}}] \\ = & - \frac{4 q^{2} d x}{4 π ε_{0} {(d^{2} - x^{2})}^{2}} . . . . . (1) \end{array}$
Now, $\begin{array}{rcl} \frac{d U}{d x} & = & 0 at x = 0 \end{array}$
Again differentiating the above equation (1), we get:
So, $\begin{array}{rcl} \frac{d^{2} U}{d x^{2}} & = & - \frac{2 d q^{2}}{4 π ε_{0}} [\frac{2}{{(d^{2} - x^{2})}^{2}} - \frac{8 x^{2}}{{(d^{2} - x^{2})}^{3}}] \end{array}$
$\begin{array}{rcl} at x & = & 0 \\ \frac{d^{2} U}{d x^{2}} & = & \frac{- 2 d q^{2}}{4 π ε_{0}} \times (\frac{2}{d^{4}} - 0) \\ = & \frac{4 q^{2}}{4 π ε_{0}} \times \frac{1}{d^{3}} < 0 \end{array}$
Hence, the charge q₂ will be in unstable equilibrium.

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