NCERT Exemplar Solutions Class 12 Physics Electrostatic Potential and Capacitance — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 12 Physics — Electrostatic Potential and Capacitance, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 2.1 to 2.13; here they are Questions 1 to 13. Questions 1 to 6 are MCQ I (exactly one option is correct) and Questions 7 to 13 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Potential and field. $\mathbf{E}$ points from high to low potential, and $V_B – V_A = -\displaystyle\int_A^B \mathbf{E}\cdot d\mathbf{l}$. The work done on a charge $q$ moved from A to B is $q(V_B-V_A)$ and does not depend on the path.
• Conductors. In equilibrium the field inside is zero, so the whole conductor is at one potential.
• Equipotentials. They are perpendicular to field lines and are closest together where the field is strongest.
• Capacitors. $Q = CV$. Whether $Q$ or $V$ stays fixed depends on whether the battery is still connected.

Key insight. Always ask what is held fixed. Connected to a battery, $V$ is fixed; once disconnected, $Q$ is fixed. Questions 1, 11 and 13 turn on that single decision.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

A capacitor of $4\ \mu\text{F}$ is connected in the circuit of Fig. 2.1. The internal resistance of the battery is $0.5\ \Omega$. The amount of charge on the capacitor plates will be

The circuit has three parallel branches between two junctions: the top branch holds the $4\ \mu$F capacitor in series with a $10\ \Omega$ resistor; the middle branch is the $2.5$ V battery; the bottom branch is a $2\ \Omega$ resistor.

Solution. In the steady state the capacitor blocks direct current, so the $10\ \Omega$ resistor in series with it carries no current and drops no voltage. The only current is the loop through the battery (with its $0.5\ \Omega$ internal resistance) and the $2\ \Omega$ resistor:

$$I = \frac{2.5}{2 + 0.5} = 1\ \text{A}.$$

The voltage across the $2\ \Omega$ resistor is $1 \times 2 = 2$ V. That is the terminal voltage of the battery, $2.5 – 1\times0.5 = 2$ V, and it is also the voltage across the capacitor branch, which lies in parallel with it. With no drop across the $10\ \Omega$ resistor, the capacitor sees the full $2$ V:

$$Q = CV = 4\ \mu\text{F} \times 2\ \text{V} = 8\ \mu\text{C}.$$

The common slip is to use the full $2.5$ V (giving $10\ \mu$C, not an option) or to answer $0$ because “no current flows”.

Question 2

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

Solution. The force on a positive charge is $q\mathbf{E}$, along the field, so a charge released from rest moves along $\mathbf{E}$. The field does positive work on it, and by energy conservation its kinetic energy rises while its potential energy falls. Uniformity of the field does not stop this: a constant force still does work as the charge moves. That rules out (a), (b) and (d).

Question 3

Figure 2.2 shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

Three panels, each with vertical dashed lines labelled $10$ V, $20$ V, $30$ V, $40$ V and $50$ V from left to right, but with different spacings. In each panel A lies on the $20$ V line and B on the $40$ V line.

Solution. The work done in moving a charge $q$ from A to B is $W = q(V_B – V_A)$. It depends only on the potentials at the end points, not on the spacing of the equipotentials or the path. In every panel A is at $20$ V and B at $40$ V, so $W = q \times 20$ V in all three. The different spacings only change how strong the field is between A and B, not the potential difference.

Question 4

The electrostatic potential on the surface of a charged conducting sphere is $100$ V. Two statements are made:

$S_1$: At any point inside the sphere, the electric intensity is zero. $S_2$: At any point inside the sphere, the electrostatic potential is $100$ V.

Which of the following is a correct statement?

Solution. Inside a conductor in equilibrium the field is zero, so $S_1$ is true. Now $V_B – V_A = -\int_A^B \mathbf{E}\cdot d\mathbf{l}$; with $\mathbf{E} = 0$ everywhere inside, the integral vanishes between any two interior points, and between an interior point and the surface. So the potential inside is constant and equal to the surface value, $100$ V: $S_2$ is true. Because $S_2$ follows directly from $S_1$, the statements are not independent.

Question 5

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

Solution. Far from the collection, its size and shape are negligible and it behaves as a single point charge equal to the net charge $Q \ne 0$. The potential $V = Q/4\pi\varepsilon_0 r$ depends only on the distance $r$, so surfaces of constant $V$ are spheres centred on the collection. (Had the net charge been zero, the dipole term would take over and the surfaces would no longer be spherical.)

Question 6

A parallel plate capacitor is made of two dielectric blocks in series. One block has thickness $d_1$ and dielectric constant $k_1$, the other thickness $d_2$ and dielectric constant $k_2$, as in Fig. 2.3. This arrangement can be thought of as a dielectric slab of thickness $d\ (= d_1 + d_2)$ and effective dielectric constant $k$. The $k$ is

The figure shows two layers stacked, one hatched one way ($k_1$) and one the other ($k_2$). Both thicknesses are labelled $d_1$ in the figure, which is a printing slip; the text gives $d_1$ and $d_2$.

Solution. Two layers of the same plate area $A$ in series: add the reciprocals of the capacitances.

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d_1}{k_1\varepsilon_0A} + \frac{d_2}{k_2\varepsilon_0A}, \qquad C = \frac{k\varepsilon_0 A}{d_1+d_2}.$$

Equating the two and solving for $k$:

$$\frac{d_1+d_2}{k} = \frac{d_1}{k_1} + \frac{d_2}{k_2} = \frac{k_2d_1 + k_1d_2}{k_1k_2} \;\Rightarrow\; k = \frac{k_1k_2(d_1+d_2)}{k_1d_2 + k_2d_1}.$$

Misprint in option (c). The book prints the denominator as $k_1d_1 + k_2d_2$. The correct denominator is $k_1d_2 + k_2d_1$, as derived above. Option (c) is the intended answer, and the key gives (c); the other options do not follow from the series derivation. As a check, when $d_1 = d_2$ the result reduces to $k = \dfrac{2k_1k_2}{k_1+k_2}$, which is option (d); that is a special case, not the general answer, so (c) is the one to choose.

Question 7

Consider a uniform electric field in the $\hat{\mathbf{z}}$ direction. The potential is a constant

Solution. For $\mathbf{E} = E\,\hat{\mathbf{z}}$, $dV = -E\,dz$, so $V = -Ez + \text{const}$. The potential changes only with $z$, so it is not constant in all space (a). At a fixed $z$ it is the same for every $x$ and every $y$, so the equipotentials are the planes $z = \text{const}$, parallel to the $x$-$y$ plane. That makes (b), (c) and (d) all true; they are three ways of saying the same thing.

Question 8

Equipotential surfaces

Solution. For neighbouring equipotentials differing by a fixed $\Delta V$ and separated by $\Delta l$, the field is $E \approx \Delta V/\Delta l$. A large $E$ therefore means a small $\Delta l$: the surfaces crowd together. That is (a).

At a conductor’s surface $E = \sigma/\varepsilon_0$, so the field is largest where the surface charge density $\sigma$ is largest, and the equipotentials crowd there. Charge density is greatest at sharp edges and points, so (b) and (c) follow. (d) fails because the field is generally not uniform (it is only for a uniform field that equal steps in $V$ are equally spaced).

Question 9

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

Solution. Along an equipotential $V$ does not change, so $W = q\,\Delta V = 0$; that is (c), and it rules out (d). Equivalently, $\mathbf{E}$ is perpendicular to the surface and so to every displacement $d\mathbf{l}$ along it, giving $\mathbf{E}\cdot d\mathbf{l} = 0$.

For (b), read the quantity as the work done per unit charge by the external agent in moving a test charge from A to B, which is by definition $-\int_A^B \mathbf{E}\cdot d\mathbf{l}$. This general definition applies on any path, including this one, where it happens to equal zero. So (a), which says it cannot be so defined, is false, and (b) is true.

Question 10

In a region of constant potential

Solution. $\mathbf{E} = -\nabla V$, so a constant $V$ means $\mathbf{E} = \mathbf{0}$ throughout the region. That is (b). Zero is a uniform field only in a trivial sense, and the question intends something non-trivial, so (a) is not the answer. By Gauss’s law $\nabla\cdot\mathbf{E} = \rho/\varepsilon_0$, so $\mathbf{E} = 0$ forces $\rho = 0$: no charge inside, which is (c).

(d) fails because of the word “necessarily”. A region inside a conductor’s cavity, or inside a conductor generally, is shielded: placing a charge outside rearranges the induced charges and leaves the field in the region at zero.

Question 11

In the circuit of Fig. 2.4, initially key $K_1$ is closed and key $K_2$ is open. Then $K_1$ is opened and $K_2$ is closed (order is important). Take $Q_1’$ and $Q_2’$ as the charges on $C_1$ and $C_2$, and $V_1$ and $V_2$ as the voltages.

The figure shows a battery of emf $E$ connected through $K_1$ across $C_1$; $C_2$ is joined to the same two nodes through $K_2$.

Solution. With $K_1$ closed, $C_1$ charges to the battery’s emf, $Q = C_1E$. Opening $K_1$ isolates $C_1$ from the battery, so $Q$ is now fixed. Closing $K_2$ connects $C_2$ to $C_1$ with the two capacitors in parallel, joined by wires that bring them to the same potential:

• (a) $V_1 = V_2$. True: the two capacitors are connected plate to plate.
• (b) $Q_1′ = Q_2’$. False in general: $Q_1′ = C_1V$ and $Q_2′ = C_2V$ are equal only if $C_1 = C_2$.
• (d) $Q_1′ + Q_2′ = Q$. True: the total charge on the connected plates is conserved.
• (c) $C_1V_1 + C_2V_2 = C_1E$. Since $Q_1′ = C_1V_1$ and $Q_2′ = C_2V_2$, the left side is $Q_1′ + Q_2’$, and the right side is $C_1E = Q$. So (c) is exactly statement (d) written in terms of capacitances and voltages. It is true.

Printed key error. The key prints (a), (d) and omits (c). But (c) is the same equation as (d), so it cannot be false when (d) is true. The correct answer is (a), (c), (d).

Question 12

If a conductor has a potential $V \neq 0$ and there are no charges anywhere else outside, then

Solution. A potential $V \ne 0$ relative to infinity with no other charges around can only come from charge on the conductor itself. So there must be charge somewhere on it: on the surface, or inside, which is (a).

In the material of the conductor, in equilibrium, $\mathbf{E} = 0$, so Gauss’s law gives zero charge in the conducting body: (b) is true. Free charge can only sit on the surface, or on the surface of a cavity inside.

(c) says charges are only on the surface. That fails because charge may sit on the wall of a cavity inside the conductor, which is why (a) says “or inside”. (d) says charges must be inside, which is false for a solid conductor, where all charge sits on the outer surface.

Question 13

A parallel plate capacitor is connected to a battery as shown in Fig. 2.5 (the capacitor $C$ in series with the battery and a key $K$). Consider two situations:

A: Key $K$ is kept closed and plates of the capacitor are moved apart using an insulating handle. B: Key $K$ is opened and plates of the capacitor are moved apart using an insulating handle.

Choose the correct option(s).

Solution. Moving the plates apart always changes the capacitance: $C = \varepsilon_0A/d$ falls as $d$ rises. What else changes depends on the battery.

• A (key closed). The battery holds the voltage at its emf, so $V$ is fixed. Since $C$ falls, $Q = CV$ falls: charge flows back into the battery. (c) is true and (a) is false.
• B (key open). The capacitor is isolated, so the charge has nowhere to go and $Q$ is fixed. With $C$ falling, $V = Q/C$ rises. (d) is true and (b) is false.

Common mistakes

• Using $2.5$ V on the capacitor in Question 1. The battery’s internal resistance drops $0.5$ V; the capacitor branch sees the terminal voltage, $2$ V.
• Thinking path or spacing matters in Question 3. Work between two points depends only on the potential difference.
• Treating (c) as different from (d) in Question 11. Both say the charge is conserved; the printed key leaves (c) out.
• Copying the printed denominator in Question 6. The correct series formula has $k_1d_2 + k_2d_1$, not $k_1d_1 + k_2d_2$.
• Calling “zero field” a uniform field in Question 10. And forgetting that shielding stops an outside charge from changing the field in a shielded region.
• Holding the wrong quantity fixed in Question 13. Connected to the battery, $V$ is fixed; disconnected, $Q$ is fixed.
• Mixing up (b) and (c) in Question 12. Charge lives on the surface, but may also sit on the wall of an internal cavity.

Practise next

• Exemplar Electrostatic Potential and Capacitance: Very Short, Short and Long Answer Questions — where the same ideas are applied to charged spheres, equipotentials, capacitor networks and forces on conductors.
• NCERT textbook Chapter 2 — Electrostatic Potential and Capacitance for the theory of potential, equipotentials, capacitors and dielectrics.
• Exemplar Electric Charges and Fields — the previous unit, from which the field and Gauss’s law ideas here are carried over.
• Exemplar Current Electricity — the next unit, which builds on the circuit reasoning used in Question 1.