NEET Electrostatic Potential and Capacitance PYQs & Practice Questions

10 free NEET Physics questions on Electrostatic Potential and Capacitance, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Electric Field of a Dipole · Dielectrics in Capacitors · Electric Potential due to a Point Charge · Equipotential Surfaces · Capacitors in Combination · Energy Stored in a Capacitor · Dielectrics and Polarization · Potential Energy in a Uniform Field · Field and Potential Inside a Conductor · Axial Potential Due to Electric Dipole (Assertion-Reason) · Equivalent Capacitance - Wheatstone-Bridge-like Network · Potential at Surface vs Nearby External Point of a Charged Shell

1. Q1NEET 2026Potential Energy and Speed Near a Charged Sphere

   Consider a fixed uniformly charged insulating sphere with radius R and total charge +Q. A point charge -q (q << Q) with mass m is released from rest at a distance of 3R from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is: (epsilon_0 is the permittivity of vacuum; neglect gravitational forces).

   • A. sqrt(2Qq/(3*pi*epsilon_0*m*R))
   • B. sqrt(Qq/(3*pi*epsilon_0*m*R))
   • C. sqrt(Qq/(4*pi*epsilon_0*m*R))
   • D. sqrt(3Qq/(4*pi*epsilon_0*m*R))
   Show answer

   Answer: A — sqrt(2Qq/(3*pi*epsilon_0*m*R))

   Outside the sphere, potential V(r) = kQ/r (k=1/4*pi*epsilon_0). Potential energy of -q at distance r is U(r) = -kQq/r. By energy conservation from 3R to R: (1/2)mv^2 = U(3R) - U(R) = -kQq/(3R) + kQq/R = kQq(2/3R) = 2kQq/(3R). Substituting k=1/(4*pi*epsilon_0): v^2 = Qq/(3*pi*epsilon_0*m*R), so v = sqrt(Qq/(3*pi*epsilon_0*m*R)).

2. Q2NEET 2026Potential due to Uniformly Charged Sphere

   A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius R, having uniform positive charge density ρ. The initial and final positions of the charge are marked by A and B at distances 2R and 3R respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is ρR²/(n∈₀). The value of n is: (∈₀ is the permittivity of vacuum)

   • A. 6
   • B. 9
   • C. 18
   • D. 2
   Show answer

   Answer: B — 9

   Both A (2R) and B (3R) lie outside the sphere (radius R), so the field there is the same as that of a point charge Q = (4/3)πR³ρ at the centre. Potential outside: V(r) = Q/(4π∈₀r) = ρR³/(3∈₀r). V(A) at r=2R: ρR²/(6∈₀). V(B) at r=3R: ρR²/(9∈₀). Work done on unit charge moving from A to B = q[V(B)-V(A)] = ρR²/(9∈₀) - ρR²/(6∈₀) = -ρR²/(18∈₀). Magnitude = ρR²/(18∈₀), so n = 18.

3. Q3NEET 2026Electric Field of a Conductor with Cavity

   A point charge Q is placed inside a cavity within a solid isolated conducting sphere. Consider points A, B and C, where the magnitude of the electric fields are E_A, E_B and E_C, respectively. Point A lies inside the cavity (off-centre, near Q). Points B and C lie outside the solid sphere, at the same distance from the centre of the solid sphere but in different directions. The correct option is:

   • A. E_A ≠ 0, E_B = E_C
   • B. E_A = 0, E_B > E_C
   • C. E_A ≠ 0, E_B < E_C
   • D. E_A = 0, E_B = E_C
   Show answer

   Answer: —

   Inside the cavity, point A is off-centre relative to charge Q, so the field there (due to Q alone, since the conductor's induced charges on the cavity wall produce no field inside the cavity other than cancelling any external influence) is nonzero: E_A ≠ 0. Outside a solid isolated conductor, the induced charge distributes itself on the outer surface such that the exterior field is exactly that of a point charge Q located at the centre, regardless of where Q sits inside the cavity (shielding effect). Since the exterior field is spherically symmetric and depends only on the radial distance from the centre, and B and C are at the same distance from the centre (just in different directions), their field magnitudes are equal: E_B = E_C.

4. Q4NEET 2026Capacitors in Series and Parallel - Energy Stored

   Three identical capacitors, P, Q and S, each of capacitance C, are connected to a battery of voltage V. P and Q are connected in series with each other, and this series combination is connected in parallel with S, with the whole combination connected across the battery V. If the energy stored in capacitor P and the total energy stored in the system are U_P and U_T, respectively, then the ratio U_P/U_T is:

   • A. 1/3
   • B. 1/2
   • C. 1/6
   • D. 2/3
   Show answer

   Answer: B — 1/2

   The P-Q series combination and S are both connected directly across the battery, so each sees the full voltage V. Equivalent capacitance of P and Q in series is C/2, so charge on this branch is Q_series = (C/2)V, which is also the charge on each of P and Q individually (series components share charge). Energy in P: U_P = Q_series²/(2C) = (CV/2)²/(2C) = CV²/8. By symmetry U_Q = CV²/8 too. Energy in S: U_S = (1/2)CV² (full voltage V across capacitance C). Total: U_T = CV²/8 + CV²/8 + CV²/2 = 3CV²/4. Ratio U_P/U_T = (CV²/8)/(3CV²/4) = 1/6.

Practise all 31 Electrostatic Potential and Capacitance questions with a timer

• Timed chapter test
• Score + weak topics
• Full NEET mocks & rank

Start free for 30 days

No payment, no card. Register in 2 minutes.

5. Q5NEET 2025Dielectrics in Capacitors

   The plates of parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness (3/8)d and d/2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1 = 1.25 K2, the value of K1 is -

   • A. 2.66
   • B. 2.33
   • C. 1.60
   • D. 1.33
   Show answer

   Answer: —

   Remaining air gap = d−3d/8−d/2 = d/8. C = ε0A/(t1/k1+t2/k2+t3/1). Setting C = 2ε0A/d and solving with k1=1.25k2 gives k1 = 2.66.

6. Q6NEET 2024Axial Potential Due to Electric Dipole (Assertion-Reason)

   Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector P of magnitude, 4×10⁻⁶ C m, is ±9×10³ V. (Take 1/4πε₀ = 9×10⁹ SI units) Reason R: V = ±2P/(4πε₀r²), where r is the distance of any axial point, situated at 2 m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:

   • A. A is true but R is false.
   • B. A is false but R is true.
   • C. Both A and R are true and R is the correct explanation of A.
   • D. Both A and R are true and R is NOT the correct explanation of A.
   Show answer

   Answer: —

   V_axial = ±kp/r² = ±(9×10⁹ × 4×10⁻⁶)/2² = ±9×10³ V, so Assertion A is true. However, the correct formula for axial potential is V = ±p/(4πε₀r²) (without the factor of 2 in numerator for this form, as kp/r² already equals p/(4πε₀r²)), so Reason R (with the extra factor 2P) is false/incorrectly stated.

7. Q7NEET 2024Equivalent Capacitance - Wheatstone-Bridge-like Network

   In the following circuit, the equivalent capacitance between terminal A and terminal B is: [Figure: Bridge network with four 2μF capacitors in a square arrangement (top two, bottom two) and one 2μF capacitor as the bridge/middle connection, between terminals A and B]

   • A. 0.5 μF
   • B. 4 μF
   • C. 2 μF
   • D. 1 μF
   Show answer

   Answer: B — 4 μF

   This is a balanced Wheatstone-bridge-like capacitor network (all capacitors equal 2μF), so the bridge (middle) capacitor carries no charge and can be removed. The top branch has two 2μF in series = 1μF, and the bottom branch has two 2μF in series = 1μF. These two 1μF are in parallel: 1+1 = 2μF.

8. Q8NEET 2024Potential at Surface vs Nearby External Point of a Charged Shell

   A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is: (Take 1/4πε₀ = 9×10⁹ SI units) [Figure: spherical shell of radius R=3cm, q=1μC, C is centre, P is a point on the shell surface]

   • A. 0.5 × 10⁵
   • B. zero
   • C. 3 × 10⁵
   • D. 1 × 10⁵
   Show answer

   Answer: A — 0.5 × 10⁵

   For a charged spherical shell, the potential is the same at the centre (C) and at the surface (any point including P on the surface), both equal to kq/R. So VP - VC = 0.

Practise all 31 Electrostatic Potential and Capacitance questions with a timer

• Timed chapter test
• Score + weak topics
• Full NEET mocks & rank

Start free for 30 days

No payment, no card. Register in 2 minutes.

9. Q9NEET 2024Parallel Plate Capacitor - Moving Plates Closer (Battery Connected)

   If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then A. the charge stored in it, increases. B. the energy stored in it, decreases. C. its capacitance increase. D. the ratio of charge to its potential remains the same. E. the product of charge and voltage increases. Choose the most appropriate answer form the options given below:

   • A. B, D and E only
   • B. A, B and C only
   • C. A, B and E only
   • D. A, C and E only
   Show answer

   Answer: C — A, B and E only

   With battery connected (constant V), moving plates closer decreases separation d, so capacitance C=ε0A/d increases (C, true). Since Q=CV and V is constant, Q increases as C increases (A, true). Energy stored = QV/2, so as Q increases with V constant, QV (and hence energy) increases (E, true, and B is false since energy increases not decreases). D is false since Q/V = C which changes (increases), not stays the same. So A, C and E only.

10. Q10NEET 2023Equivalent Capacitance - Series-Parallel Combination

    The equivalent capacitance of the system shown in the following circuit is: [Circuit shows a 3μF capacitor from A in series with a parallel combination of two 3μF capacitors, connecting to B]

    • A. 3 μF
    • B. 6 μF
    • C. 9 μF
    • D. 2 μF
    Show answer

    Answer: C — 9 μF

    The two 3μF capacitors in parallel combine to give 6μF. This 6μF is then in series with the first 3μF capacitor: CAB = (3×6)/(3+6) = 18/9 = 2μF.