Electrostatic Potential and Capacitance NEET PYQs (2005–2026) with Solutions

10 solved Electrostatic Potential and Capacitance PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumEnergy stored in capacitor combinations

   Three identical capacitors, $P$, $Q$ and $S$, each of the capacitance $C$, are connected to a battery of voltage $V$, as shown in the figure. If the energy stored in the capacitor $P$ and total energy stored in the system are $U_P$ and $U_T$, respectively, then the ratio $\frac{U_P}{U_T}$ is:

   1. AOption A: $2/3$
   2. BOption B: $1/3$
   3. COption C: $1/2$
   4. DOption D: $1/6$

   Show answer & explanationHide answer & explanation

   Correct answer: (D) $1/6$

   Explanation

   From the figure, $P$ and $Q$ are in series, and this pair is in parallel with $S$, all across the battery $V$. In the series pair each capacitor gets $\frac{V}{2}$, so $U_P = \frac{1}{2}C\left(\frac{V}{2}\right)^2 = \frac{CV^2}{8}$. Equivalent capacitance $C_{eq} = \frac{C}{2} + C = \frac{3C}{2}$, so $U_T = \frac{1}{2}\cdot\frac{3C}{2}V^2 = \frac{3CV^2}{4}$. $\frac{U_P}{U_T} = \frac{1/8}{3/4} = \frac{1}{6}$, option (D).

2. Question 2 (NEET 2026)

   NEET 2026MediumCapacitors & Capacitance

   Five capacitors of capacitances C₁ = C₂ = C₃ = C₄ = 10 μF and C₅ = 2.5 μF are connected as shown, along with a battery of 50 V. The equivalent capacitance and the charges on each capacitor respectively are:

   1. AOption A: 5 μF, 125 μC on C₁ to C₄ and 25 μC on C₅
   2. BOption B: 5 μF, 125 μC on all capacitors
   3. COption C: 5 μF, 250 μC on all capacitors
   4. DOption D: 4 μF, 250 μC on C₁ to C₄ and 125 μC on C₅

   Show answer & explanationHide answer & explanation

   Correct answer: (B) 5 μF, 125 μC on all capacitors

   Explanation

   Capacitors C₁, C₂, C₃ and C₄ (each 10 μF) are connected in series. Equivalent capacitance of series combination: 1/C = 1/10 + 1/10 + 1/10 + 1/10 ⇒ C = 2.5 μF This 2.5 μF combination is in parallel with C₅ = 2.5 μF. Therefore: Ceq = 2.5 + 2.5 = 5 μF Charge on equivalent branch: Q = CV = 2.5 × 50 Q = 125 μC For capacitors in series, charge remains same: q₁ = q₂ = q₃ = q₄ = 125 μC For capacitor C₅: q₅ = 2.5 × 50 = 125 μC Hence all capacitors carry 125 μC and equivalent capacitance is 5 μF. Therefore option (B) is correct.

3. Question 3 (NEET 2026)

   NEET 2026MediumPotential due to a 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 $\rho$, as shown in the figure. The initial and final positions of the charge are marked by $A$ and $B$ at distance $2R$ and $3R$ respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is $\frac{\rho R^2}{n\varepsilon_0}$. The value of $n$ is : ($\varepsilon_0$ is the permittivity of vacuum)

   1. AOption A: $2$
   2. BOption B: $6$
   3. COption C: $9$
   4. DOption D: $18$

   Show answer & explanationHide answer & explanation

   Correct answer: (D) $18$

   Explanation

   The electrostatic force is conservative, so the work depends only on the end points $A$ and $B$, not on the path through the sphere. Both $A$ and $B$ are outside the sphere, where it acts like a point charge $Q = \frac{4}{3}\pi R^3\rho$ at its centre: $V(r) = \frac{Q}{4\pi\varepsilon_0 r} = \frac{\rho R^3}{3\varepsilon_0 r}$. $V_A - V_B = \frac{\rho R^3}{3\varepsilon_0}\left(\frac{1}{2R} - \frac{1}{3R}\right) = \frac{\rho R^3}{3\varepsilon_0}\cdot\frac{1}{6R} = \frac{\rho R^2}{18\varepsilon_0}$. For a unit charge, the magnitude of the work done (by the field, or equally by the agent moving it slowly) is $|V_A - V_B| = \frac{\rho R^2}{18\varepsilon_0}$, so $n = 18$, option (D).

4. Question 4 (NEET 2026)

   NEET 2026MediumEnergy Stored in Capacitors

   Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

   1. AOption A: 0.5 × 10⁻⁶ J
   2. BOption B: 1.0 J
   3. COption C: 1.0 × 10⁻⁶ J
   4. DOption D: 0.5 J

   Show answer & explanationHide answer & explanation

   Correct answer: (A) 0.5 × 10⁻⁶ J

   Explanation

   Energy lost = (1/2)[C1C2/(C1+C2)]V². For C1=C2=200 pF and V=100 V: Energy lost = (1/2)(200×200/400)×10⁻¹²×100² = 0.5×10⁻⁶ J. Hence option (A) is correct.

5. Question 5 (NEET 2026)

   NEET 2026MediumElectrostatics of conductors with a cavity

   A point charge $Q$ is placed inside a cavity within a solid isolated conducting sphere. Consider points $A$, $B$ and $C$ as shown in the figure, where the magnitudes of the electric fields are $E_A$, $E_B$, $E_C$, respectively. The points $B$ and $C$ are at the same distance from the center of the solid sphere. The correct option is :

   1. AOption A: $E_A = 0, E_B = E_C$
   2. BOption B: $E_A \neq 0, E_B = E_C$
   3. COption C: $E_A = 0, E_B > E_C$
   4. DOption D: $E_A \neq 0, E_B < E_C$

   Show answer & explanationHide answer & explanation

   Correct answer: (B) $E_A \neq 0, E_B = E_C$

   Explanation

   Point $A$ is inside the cavity, near the charge $Q$. In the cavity the field of $Q$ (together with the induced charge $-Q$ on the cavity wall) is not zero, so $E_A \neq 0$. A charge $+Q$ is induced on the outer surface of the isolated sphere. The conductor shields the outside from the cavity, so this outer charge spreads uniformly over the spherical surface no matter where $Q$ sits inside the cavity. Outside, the field is therefore that of a point charge $Q$ at the sphere's centre: $E = \frac{Q}{4\pi\varepsilon_0 r^2}$. $B$ and $C$ are at the same $r$, so $E_B = E_C$. Hence option (B).

6. Question 6 (NEET 2026)

   NEET 2026MediumElectrostatic potential energy and energy conservation

   Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+Q$. A point charge $-q$ ($q \ll Q$) 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: ($\varepsilon_0$ is the permittivity of vacuum, neglect gravitational forces).

   1. AOption A: $\sqrt{\frac{3Qq}{4\pi\varepsilon_0 mR}}$
   2. BOption B: $\sqrt{\frac{2Qq}{3\pi\varepsilon_0 mR}}$
   3. COption C: $\sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$
   4. DOption D: $\sqrt{\frac{Qq}{4\pi\varepsilon_0 mR}}$

   Show answer & explanationHide answer & explanation

   Correct answer: (C) $\sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$

   Explanation

   The charge $-q$ is attracted towards the sphere, so its potential energy $U = -\frac{Qq}{4\pi\varepsilon_0 r}$ decreases and its kinetic energy increases. Outside and on the sphere it behaves like a point charge $Q$ at the centre. Energy conservation: $\frac{1}{2}mv^2 = U(3R) - U(R) = \frac{Qq}{4\pi\varepsilon_0}\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{Qq}{4\pi\varepsilon_0}\cdot\frac{2}{3R} = \frac{Qq}{6\pi\varepsilon_0 R}$. $v^2 = \frac{Qq}{3\pi\varepsilon_0 mR} \Rightarrow v = \sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$, option (C).

7. Question 7 (NEET 2026)

   NEET 2026MediumConductors & Dielectrics

   Which of the following statements are correct? A. Inside a conductor, the electrostatic field is zero. B. Electric field at the surface of a charged conductor does not depend on its surface charge density. C. The interior of a charged conductor can have no excess charge in the static situation. D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point. E. The electrostatic potential is zero everywhere inside a charged conductor. Choose the correct answer from the options given below:

   1. AOption A: A, B and D only
   2. BOption B: A, C and E only
   3. COption C: A, C and D only
   4. DOption D: C, D and E only

   Show answer & explanationHide answer & explanation

   Correct answer: (C) A, C and D only

   Explanation

   A is true: electric field inside a conductor is zero. B is false: E=σ/ε₀ depends on surface charge density. C is true: excess charge resides only on surface. D is true: electric field is normal to surface. E is false: potential inside conductor is constant, not necessarily zero. Hence option (C) is correct.

8. Question 8 (NEET 2025)

   NEET 2025HardCapacitors & Capacitance

   The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant $K_1$ and $K_2$ with thicknesses 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 $K_1$ = 1.25 $K_2$, the value of $K_1$ is :

   1. AOption A: 2.66
   2. BOption B: 2.33
   3. COption C: 1.60
   4. DOption D: 1.33

   Show answer & explanationHide answer & explanation

   Correct answer: (A) 2.66

   Explanation

   Using the equivalent separation formula for dielectric slabs in series: d_eff = (3d/8K1) + (d/2K2) + d/8. Since the new capacitance is twice the original, d_eff = d/2. Using K1 = 1.25K2 and solving gives K1 ≈ 2.66.

9. Question 9 (NEET 2024)

   NEET 2024MediumCapacitors & Capacitance

   In the following circuit, the equivalent capacitance between terminal A and terminal B is:

   1. AOption A: 0.5 μF
   2. BOption B: 4 μF
   3. COption C: 2 μF
   4. DOption D: 1 μF

   Show answer & explanationHide answer & explanation

   Correct answer: (C) 2 μF

   Explanation

   The circuit is a balanced bridge because all capacitances are equal. Hence, no potential difference exists across the middle capacitor, so it can be ignored. Top branch: $$C_t = \frac{2 \times 2}{2+2} = 1\,\mu F$$ Bottom branch: $$C_b = 1\,\mu F$$ These two branches are in parallel: $$C_{eq} = 1 + 1 = 2\,\mu F$$ Hence, option (C) is correct.

10. Question 10 (NEET 2024)

    NEET 2024MediumElectrostatic Potential

    A thin spherical shell is charged by some source. The potential difference between two points C and P (in V) shown in the figure is: (Take 1/(4πϵ₀) = 9 × 10⁹ SI units)

    1. AOption A: 0.5 × 10⁵
    2. BOption B: zero
    3. COption C: 3 × 10⁵
    4. DOption D: 1 × 10⁵

    Show answer & explanationHide answer & explanation

    Correct answer: (B) zero

    Explanation

    For a charged conducting spherical shell, the electric potential is constant throughout the interior and on the surface. Since both points C and P lie inside/on the shell, they are at the same potential. Therefore: $$V_C - V_P = 0$$ Hence, option (B) is correct.