Previous year questions: Electrostatic Potential and Capacitance

Ten questions follow, taken from official CBSE papers in the collected corpus. Each item gives the paper label and marks. Attempt the question before reading the answer, and state clearly in every dielectric question which quantity is fixed.

Q1 (CBSE Board 2023, 5 marks)

Define electric potential at a point and write its SI unit. Derive an expression for the equivalent capacitance of two capacitors joined in series. Charges +q and -q are located at (3a, 0) and (0, 4a), and a third charge Q is placed at the origin. Find Q in terms of q so that the potential energy of the system is zero.

Answer: The potential at a point is the work per unit positive charge needed to bring the charge there from infinity, and its SI unit is the volt. For two capacitors C1 and C2 in series, both carry charge Q and the potential differences add: V = Q/C1 + Q/C2 = Q(1/C1 + 1/C2), so 1/C = 1/C1 + 1/C2. For the charges, the pair distances are 3a, 4a and 5a. The total energy is k times [Qq/(3a) - Qq/(4a) - q^2/(5a)]. Set this to zero, divide by kq/a, and get Q/3 - Q/4 - q/5 = 0. The first two terms give Q/12, so Q/12 = q/5 and Q = 12q/5.

Q2 (CBSE Board 2024, 1 mark)

Ten capacitors, each of capacitance 1 microfarad, are connected in parallel to a source of 100 V. What is the total energy stored in the system?

Answer: Capacitors in parallel add, so the equivalent capacitance is 10 microfarad. Energy stored is (1/2) C V^2 = 0.5 × 10 × 10^-6 × (100)^2 = 0.05 J, which is 5 × 10^-2 J. The full 100 V appears across each capacitor, and the total energy is also ten times the energy of one capacitor, which gives the same answer.

Q3 (CBSE Board 2024, 3 marks)

An air-filled parallel plate capacitor with plate separation 1 mm has a capacitance of 20 pF. It is charged to 4.0 microcoulomb. Calculate the work done to pull its plates apart to a separation of 5 mm, assuming the charge stays the same.

Answer: At constant charge the work done equals the increase in stored energy, U = Q^2 / 2C. The new capacitance at five times the separation is one fifth of 20 pF, which is 4 pF. The initial energy is (4 × 10^-6)^2 / (2 × 20 × 10^-12) = 0.4 J. The final energy is (4 × 10^-6)^2 / (2 × 4 × 10^-12) = 2.0 J. The work done is 2.0 - 0.4 = 1.6 J, and the external agent does this positive work.

Q4 (CBSE Board 2024, 5 marks)

Derive the expression for the capacitance of a parallel plate capacitor with a dielectric medium between the plates. A charge of 6 microcoulomb is given to a hollow metallic sphere of radius 0.2 m. Find the potential at its surface and at its centre.

Answer: With plate area A, separation d and dielectric constant K, the field in the dielectric is E = Q / (K epsilon0 A). The potential difference is V = E d = Q d / (K epsilon0 A), so C = K epsilon0 A / d. For a conducting sphere of radius 0.2 m carrying 6 microcoulomb, the potential at the surface is k Q / R = 9 × 10^9 × 6 × 10^-6 / 0.2 = 2.7 × 10^5 V. Inside a conductor the field is zero, so the potential is constant throughout, and the potential at the centre equals the surface value, 2.7 × 10^5 V. A useful check: the potential of a conducting sphere is fixed by its radius and charge, so two spheres of different radii at the same potential carry charges in proportion to their radii.

Q5 (CBSE Board 2025, 1 mark)

A metal sheet is inserted between the plates of a parallel plate capacitor of capacitance C so that it partly occupies the space between the plates. The capacitance

Answer: The capacitance becomes greater than C. A metal sheet of thickness t removes a region of thickness t from the field, so the effective separation becomes d - t, and the capacitance becomes epsilon0 A / (d - t), which is larger than epsilon0 A / d. The metal sheet does not change the field in the air gaps, and the answer stays greater than C for any sheet thinner than the separation.

Q6 (CBSE Board 2026, 3 marks)

A parallel plate capacitor of capacitance C has a dielectric slab between its plates. It is charged to a potential difference V by connecting it across a battery. The battery is then disconnected. If the dielectric slab is withdrawn from the capacitor, how are the capacitance, the electric field between the plates and the stored energy affected? Justify each answer.

Answer: With the battery disconnected the charge Q stays fixed. When the slab is withdrawn the capacitance falls to C / K, because the dielectric was multiplying it by K. The field between the plates rises to K times its value inside the dielectric, because the charge density on the plates is unchanged and the field in air is Q / (epsilon0 A). The stored energy becomes Q^2 / 2C with the smaller capacitance, so it rises by the factor K. The external agent does positive work in pulling the slab out.

Q7 (CBSE Board 2026, 5 marks)

Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air between the plates. Two air-filled capacitors of capacitances C1 and C2 are connected in parallel across a dc battery, and a slab of dielectric constant K is then inserted between the plates of C2. Explain how the total capacitance, total charge and total stored energy change.

Answer: For the air capacitor, the field is E = Q / (epsilon0 A) and the potential difference is E d, so C = epsilon0 A / d. With the battery connected, the potential difference across each capacitor stays equal to the battery voltage. Inserting the slab changes C2 to K C2, so the total capacitance becomes C1 + K C2, which is larger. The total charge becomes V (C1 + K C2), also larger. The total energy is (1/2)(C1 + K C2) V^2, which rises too. The battery supplies the extra charge and the extra energy. If the slab were inserted after disconnecting the battery instead, the charge would stay fixed while the potential fell, and the stored energy would fall by K, so read the question for the battery condition first.

Q8 (CBSE Board 2024, 1 mark)

The capacitance of a parallel plate capacitor with a medium of dielectric constant K = 4 between the plates is C. If this medium is removed, the capacitance of the capacitor becomes

Answer: C / 4. The dielectric multiplies the capacitance by its dielectric constant, so removing it divides the capacitance by K. With K = 4, the new capacitance is one quarter of the original value. The air-filled capacitor has the smaller capacitance because the field between its plates is stronger for the same charge.

Q9 (CBSE SQP 2021-22, 1 mark)

Three capacitors of 2 microfarad, 3 microfarad and 6 microfarad are joined in series with each other. The equivalent capacitance is

Answer: 1 microfarad. The reciprocals add: 1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1 per microfarad. Taking the reciprocal gives C = 1 microfarad. As a check, a series equivalent is always smaller than the smallest individual capacitor, and 1 is smaller than 2.

Q10 (CBSE Additional PQ 2022-23, 1 mark)

A parallel plate capacitor of capacitance C0 is connected to a battery of voltage V, so the capacitor charges fully. A slab of dielectric material of constant K is then introduced between the plates while the battery stays connected. How do the charge stored and the energy stored change?

Answer: With the battery connected the potential difference stays V. The capacitance becomes K times larger, so the charge Q = C V becomes K times larger. The stored energy is (1/2) C V^2, so it also becomes K times larger. The battery supplies both the extra charge and the extra energy.