Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 12.
Q2079 (2026)
A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates.
A sheet having thickness $\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :
- $\frac{3\mathrm{KC}}{2\mathrm{K} + 1}$
- $\frac{\mathrm{CK}}{2 + \mathrm{K}}$
- $\frac{3\mathrm{CK}^2}{(2\mathrm{K} + 1)^2}$
- $\frac{4\mathrm{KC}}{3\mathrm{K} - 1}$
### Related Formula
$$C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$$
Alternatively, treat as two capacitors in series: $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$
### Core Logic
Initial capacitance (vacuum): $C = \frac{A \epsilon_{0}}{d}$.
{{SOL_IMG1}}
The system can be modelled as two capacitors in series:
1. A vacuum capacitor of thickness $d - \frac{d}{3} = \frac{2d}{3}$
2. A dielectric capacitor of thickness $\frac{d}{3}$ and dielectric constant $K$.
{{SOL_IMG2}}
{{SOL_IMG3}}
### Step 1: Capacitors in Series
The capacitances are:
$$C_1 = \frac{\epsilon_0 A}{\left(\frac{2d}{3}\right)} = \frac{3}{2} \left(\frac{\epsilon_0 A}{d}\right) = \frac{3}{2} C$$
$$C_2 = \frac{K \epsilon_0 A}{\left(\frac{d}{3}\right)} = 3K \left(\frac{\epsilon_0 A}{d}\right) = 3KC$$
Now, equivalent capacitance in series:
$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{\left(\frac{3}{2} C\right) \times (3KC)}{\frac{3}{2} C + 3KC}$$
$$C_{\text{eq}} = \frac{\frac{9}{2}KC^2}{\frac{3}{2}C(1 + 2K)}$$
$$C_{\text{eq}} = \frac{3KC}{2K + 1}$$
### Pattern Recognition
Partial dielectric filling of thickness $t$: Use formula $C_{\text{new}} = \frac{\epsilon_0 A}{d - t + t/K}$. Plugging $t = d/3$ directly gives $\frac{\epsilon_0 A}{d - d/3 + d/3K} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{3K}} = \frac{3K \epsilon_0 A}{2Kd + d} = \frac{3KC}{2K+1}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q45 (2026)
A point charge of $10^{-8}$ C is placed at origin. The work done in moving a point charge 2 $\mu$C from point A(4, 4, 2) m to B(2, 2, 1) m is ____ J. $\left(\frac{1}{4\pi\epsilon_{0}}=9\times10^{9}\text{ in SI units}\right)$
- $45 \times 10^{-6}$
- $0$
- $30 \times 10^{-6}$
- $15 \times 10^{-6}$
### Related Formula
$$W_{\text{ext}} = \Delta U = U_f - U_i$$
$$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$$
### Core Logic
Work done by external agent: $W_{\text{ext}} = \Delta U$, where $\Delta U$ is the change in potential energy.
$$W_{\text{ext}} = \frac{1}{4\pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{f}} - \frac{1}{4\pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{i}}$$
Calculate the distances of points A and B from the origin:
$r_i = |A| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6\text{ m}$
$r_f = |B| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\text{ m}$
### Step 1: Calculate Work Done
$$W_{\text{ext}} = (9 \times 10^9) \times (10^{-8} \times 2 \times 10^{-6}) \left[ \frac{1}{3} - \frac{1}{6} \right]$$
$$W_{\text{ext}} = 18 \times 10^{-5} \times \left(\frac{2-1}{6}\right)$$
$$W_{\text{ext}} = 18 \times 10^{-5} \times \frac{1}{6} = 3 \times 10^{-5}\text{ J} = 30 \times 10^{-6}\text{ J}$$
### Pattern Recognition
Electric field is conservative. Work done simply equals change in $kqq/r$ from initial to final radial coordinate. No path dependence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q26 (2026)
Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is:
(Take $k=\frac{1}{4\pi\epsilon_{0}}$ and assume $kQ^{2}>Gm^{2}$ where G is the Gravitational constant)
- $\sqrt{\frac{kQ^{2}}{4mR}\left(1-\frac{Gm^{2}}{kQ^{2}}\right)}$
- $\sqrt{\frac{kQ^{2}}{4mR}\left(1+\frac{Gm^{2}}{kQ^{2}}\right)}$
- $\sqrt{\frac{kQ^{2}}{2mR}\left(1-\frac{Gm^{2}}{kQ^{2}}\right)}$
- $\sqrt{\frac{kQ^{2}}{2mR}\left(1-\frac{Gm^{2}}{2kQ^{2}}\right)}$
### Related Formula
$$K_i + U_i = K_f + U_f$$
$$U = \frac{kq_1q_2}{r} - \frac{Gm_1m_2}{r}$$
### Core Logic
Using energy conservation from the initial state (separation $4R$) to the final state (just touching, so center-to-center separation is $2R$). Both spheres have mass $m$ and speed $u$.
Initial Energy:
$$E_i = 2\left(\frac{1}{2}mu^2\right) - \frac{Gm^2}{4R} + \frac{kQ^2}{4R}$$
Final Energy (just touching implies final velocity is zero):
$$E_f = - \frac{Gm^2}{2R} + \frac{kQ^2}{2R}$$
### Step 1: Equating Energies
$$mu^2 - \frac{Gm^2}{4R} + \frac{kQ^2}{4R} = - \frac{Gm^2}{2R} + \frac{kQ^2}{2R}$$
$$mu^2 = \left(\frac{kQ^2}{2R} - \frac{kQ^2}{4R}\right) - \left(\frac{Gm^2}{2R} - \frac{Gm^2}{4R}\right)$$
$$mu^2 = \frac{kQ^2}{4R} - \frac{Gm^2}{4R}$$
$$u^2 = \frac{1}{4mR}(kQ^2 - Gm^2)$$
### Step 2: Final Conclusion
$$u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}$$
### Pattern Recognition
When dealing with two forces (electrostatic repulsion and gravitational attraction), their potentials simply superimpose linearly. The change in total potential energy equals the loss in kinetic energy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Class 11 Physics: Gravitation
Q27 (2026)
The charge stored by the capacitor C in the given circuit in the steady state is ________ $\mu\text{C}$.
{{IMG1}}
- $12.5$
- $10$
- $7.5$
- $5$
### Related Formula
$Q = C V_c$
where $V_c$ is the steady-state voltage across the capacitor.
### Core Logic
In steady state, the capacitor acts as an open circuit (blocks DC current). We must analyze the active branches.
{{SOL_IMG1}}
The branch with the reversed-biased diode will carry no current. Current flows through the outer loop via the forward-biased diode.
Total active resistance $R_{\text{eq}} = 1\,\Omega + 4\,\Omega = 5\,\Omega$.
Current in the steady state:
$$i = \frac{V}{R_{\text{eq}}} = \frac{2.5}{5} = 0.5 \text{ A}$$
### Step 1: Voltage Calculation
The voltage across the capacitor $V_c$ is equal to the voltage drop across the $4\,\Omega$ resistor, because the branch is connected in parallel.
$$V_c = i \times 4\,\Omega$$
$$V_c = 0.5 \times 4 = 2 \text{ V}$$
### Step 2: Final Conclusion
The charge stored is:
$Q = C V_c$
$$Q = 5\,\mu\text{F} \times 2\text{ V} = 10\,\mu\text{C}$$
### Pattern Recognition
Capacitor in steady-state DC = open wire. Analyze only the paths where current can physically flow, check diode polarity, find nodal voltage across the capacitor terminals, apply Q=CV.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Class 12 Physics: Current Electricity
Class 12 Physics: Semiconductor Electronics
Q28 (2026)
Six point charges are kept $60^{circ}$ apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the centre of the circle is \_\_\_\_. ($\epsilon_{o}$ is permittivity of free space)
{{IMG1}}
- $-\frac{5Q}{8\pi\epsilon_{0}R^{2}}(\hat{\mathrm{i}}+\sqrt{3}\hat{\mathrm{j}})$
- $-\frac{Q}{4\pi\epsilon_{0}R^{2}}(\sqrt{3}\hat{\mathrm{i}}-\hat{\mathrm{j}})$
- $-\left(\frac{5Q}{8\pi\epsilon_{0}R^{2}}\right)(\hat{\mathrm{i}}-3\hat{\mathrm{j}})$
- $\frac{Q}{4\pi\epsilon_{0}R^{2}}(\sqrt{3}\hat{\mathrm{i}}-\hat{\mathrm{j}})$
### Related Formula
$$E = \frac{kQ}{r^2}$$
### Core Logic
{{SOL_IMG1}}
By symmetry and vector addition of electric fields due to the point charges placed on the circle:
$$\vec{E}_{\text{net}} = 2E_0 \cos 30^{\circ}(-\hat{i}) + 2E_0 \sin 30^{\circ}(\hat{j})$$
$$= \frac{2kQ}{r^2}\left[\frac{\sqrt{3}}{2}(-\hat{i}) + \frac{1}{2}\hat{j}\right]$$
$$= -\frac{Q}{4\pi\epsilon_0 R^2}(\sqrt{3}\hat{i} - \hat{j})$$
### Pattern Recognition
Sees: Symmetrical charge distribution on a circle.
Shortcut: Resolve components symmetrically and combine vector contributions.
Check: Matches option (2). ✓
### Chapter Mix
Class 12 Physics: Electrostatics
