Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 5.
Q14 (2025)
The electrostatic potential on the surface of uniformly charged spherical shell of radius $R = 10\mathrm{~cm}$ is $120\mathrm{~V}$. The potential at the centre of shell, at a distance $r = 5\mathrm{~cm}$ from centre, and at a distance $r = 15\mathrm{~cm}$ from the centre of the shell respectively, are:
- $120\mathrm{~V}, 120\mathrm{~V}, 80\mathrm{~V}$
- $40\mathrm{~V}, 40\mathrm{~V}, 80\mathrm{~V}$
- $0\mathrm{~V}, 0\mathrm{~V}, 80\mathrm{~V}$
- $0\mathrm{~V}, 120\mathrm{~V}, 40\mathrm{~V}$
### Related Formula
For a uniformly charged spherical shell of radius $R$ and charge $Q$:
- Inside and on the surface of the shell ($r \le R$):
$$V_{\text{in}} = V_{\text{surface}} = \frac{kQ}{R}$$
- Outside the shell ($r > R$):
$$V_{\text{out}} = \frac{kQ}{r} = V_{\text{surface}} \left(\frac{R}{r}\right)$$
### Core Logic
Let's calculate the potentials at the specified positions:
- Given surface potential at $R = 10\mathrm{~cm}$ is $120\mathrm{~V}$.
1. **At the center ($r = 0$)**:
Since the center lies inside the shell ($0 < 10\mathrm{~cm}$), the potential equals the surface potential:
$$V_{\text{centre}} = 120\mathrm{~V}$$
2. **At $r = 5\mathrm{~cm}$**:
Since $5\mathrm{~cm}$ is also inside the shell ($5 < 10\mathrm{~cm}$), the potential remains constant at the surface value:
$$V_{r=5} = 120\mathrm{~V}$$
3. **At $r = 15\mathrm{~cm}$**:
Since $15\mathrm{~cm}$ is outside the shell ($15 > 10\mathrm{~cm}$), the potential decreases inversely with distance:
$$V_{r=15} = V_{\text{surface}} \left(\frac{R}{r}\right) = 120 \times \frac{10}{15} = 80\mathrm{~V}$$
Therefore, the potentials are $120\mathrm{~V}$, $120\mathrm{~V}$, and $80\mathrm{~V}$ respectively.
### Pattern Recognition
The electric field inside a uniformly charged conducting spherical shell is zero, meaning that no work is done moving a charge inside it. Consequently, the potential remains absolutely uniform/constant from the surface all the way to the center!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q7 (2025)
A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to 'O' is 'E' is magnitude. What would be the magnitude of electric field at 'O' due to arc ABC? {{IMG1}}
- $2E$
- $\sqrt{2}E$
- $E/2$
- Zero
### Related Formula
The electric field due to a circular arc subtending an angle $\phi$ at the center is given by:
$$E_{\text{arc}} = \frac{2k\lambda}{R} \sin\left(\frac{\phi}{2}\right)$$
### Core Logic
Arc AB subtends $90^\circ$ (one quadrant) at the center. The electric field due to it is given as $E$.
Arc ABC consists of two independent quadrants: arc AB and arc BC.
Each quadrant independently creates an electric field of magnitude $E$ pointing along the bisector of that specific quadrant.
### Step 1: Vector Addition
The electric field $\vec{E}_{AB}$ is directed at $45^\circ$ away from both axes into the third quadrant.
The electric field $\vec{E}_{BC}$ is directed at $45^\circ$ towards the matching opposite quadrant.
Since $\vec{E}_{AB}$ and $\vec{E}_{BC}$ are perpendicular to each other, their resultant magnitude is:
$$E_{\text{net}} = \sqrt{E^2 + E^2} = \sqrt{2}E$$ {{SOL_IMG1}} {{SOL_IMG2}}
### Pattern Recognition
Symmetric components of a ring create orthogonal vector fields. Each $90^\circ$ arc produces a field of magnitude $E$ directed along its angular bisector. Two adjacent quadrants have bisectors separated by $90^\circ$, hence use orthogonal vector addition.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q10 (2025)
Three parallel plate capacitors $C_{1}$, $C_{2}$ and $C_{3}$ each of capacitance $5\ \mu\text{F}$ are connected as shown in figure. The effective capacitance between points A and B, when the space between the parallel plates of $C_{1}$ capacitor is filled with a dielectric medium having dielectric constant of 4, is: {{IMG1}}
- $22.5\ \mu\text{F}$
- $7.5\ \mu\text{F}$
- $9\ \mu\text{F}$
- $30\ \mu\text{F}$
### Related Formula
Capacitance modification by dielectric:
$C' = K \cdot C$
Series combination:
$$C_{\text{series}} = \frac{C_a C_b}{C_a + C_b}$$
Parallel combination:
$$C_{\text{parallel}} = C_1 + C_2$$
### Core Logic
Initial capacitance value $C = 5\ \mu\text{F}$ for all.
After dielectric insertion into $C_1$, its value becomes:
$$C_1 = 4 \times 5 = 20\ \mu\text{F}$$
The values for the others remain constant:
$$C_2 = 5\ \mu\text{F}, \quad C_3 = 5\ \mu\text{F}$$
### Step 1: Circuit Topology Analysis
From the network layout, $C_1$ and $C_2$ are configured in a series arm, which is collectively in parallel with $C_3$.
Equivalent of the series arm:
$$C_{12} = \frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \mu\text{F}$$
Adding the parallel branch $C_3$:
$$C_{\text{eq}} = C_{12} + C_3 = 4 + 5 = 9\ \mu\text{F}$$
### Pattern Recognition
Identify layout components systematically. Series components simplify via product-over-sum, then combine linearly with parallel components.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q918 (2025)
Two infinite identical charged sheets and a charged spherical body of charge density '$
ho$' are arranged as shown in figure. Then the correct relation between the electrical fields at A, B, C and D points is :
{{IMG1}}
- $\vec{E}_{A}=\vec{E}_{B}; \vec{E}_{C}=\vec{E}_{D}$
- $\vec{E}_{A}>\vec{E}_{B}; \vec{E}_{C}=\vec{E}_{D}$
- $\vec{E}_{C} \neq \vec{E}_{D}; \vec{E}_{A} > \vec{E}_{B}$
- $|\vec{E}_{A}|=|\vec{E}_{B}|; \vec{E}_{C}>\vec{E}_{D}$
### Related Formula
Principle of superposition for electric fields:
$$\vec{E}_{\text{net}} = \vec{E}_{\text{sheets}} + \vec{E}_{\text{sphere}}$$
Electric field due to an infinite uniformly charged sheet:
$$E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}$$
Electric field outside a uniformly charged spherical body:
$$E_{\text{sphere}} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} \propto \frac{1}{r^2}$$
### Core Logic
* **Between the sheets (Points $A$ and $B$):** The two identical infinite parallel sheets produce equal and oppositely directed electric fields that cancel completely ($\vec{E}_{\text{sheets}} = \vec{0}$). Thus, the net electric field is determined entirely by the charged sphere. Because point $A$ is located closer to the center of the sphere than point $B$ ($r_A < r_B$):
$$E_A > E_B \implies \vec{E}_A > \vec{E}_B$$
* **Outside the sheets (Points $C$ and $D$):** The planar electric fields reinforce each other in the same direction ($E = \frac{\sigma}{\varepsilon_0}$), while the sphere contributes a radial field. Because points $C$ and $D$ lie at different radial distances from the center of the sphere ($r_D < r_C$), the net field magnitudes differ:
$$E_C \neq E_D \implies \vec{E}_C \neq \vec{E}_D$$
Therefore, the valid relation is $\vec{E}_C \neq \vec{E}_D; \vec{E}_A > \vec{E}_B$.
### Pattern Recognition
Between two identical positively charged parallel sheets, the sheet fields cancel out completely, making the sphere the sole field source. Outside both sheets, the sheet fields reinforce each other, and asymmetric radial distances to the sphere guarantee unequal net field values.
### Evaluation Rubric / Model Answer
Option C: $\vec{E}_{C} \neq \vec{E}_{D}; \vec{E}_{A} > \vec{E}_{B}$
### Chapter Mix
Class 12 Physics: Electrostatics
Q919 (2025)
Two small spherical balls of mass 10g each with charges $-2mumathrm{C}$ and $2mumathrm{C}$, are attached to two ends of very light rigid rod of length 20 cm. The arrangement is now placed near an infinite non-conducting charge sheet with uniform charge density of $100mumathrm{C/m}^{2}$ such that length of rod makes an angle of $30^{\circ}$ with electric field generated by charge sheet. Net torque acting on the rod is:
(Take $\epsilon*{o} = 8.85\times10^{-12}\mathrm{C}^{2}/\mathrm{Nm}^{2}$)
- 112 Nm
- 1.12 Nm
- 2.24 Nm
- 11.2 Nm
### Related Formula
Electric field due to an infinite non-conducting sheet:
$$E = \frac{\sigma}{2\varepsilon_0}$$
Torque on an electric dipole:
$$\tau = pE\sin\theta$$
where $p = qd$ is the magnitude of the electric dipole moment.
### Core Logic
Given parameters:
* Charge magnitude, $q = 2\ \mu\text{C} = 2 \times 10^{-6}\text{ C}$
* Separation length, $d = 20\text{ cm} = 0.2\text{ m}$
* Surface charge density, $\sigma = 100\ \mu\text{C/m}^2 = 100 \times 10^{-6}\text{ C/m}^2$
* Orientation angle with field, $\theta = 30^{\circ}$
* Permittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/(\text{N}\cdot\text{m}^2)$
### Step 1: Compute Electric Field and Net Torque
The electric field generated by the infinite non-conducting sheet is uniform:
$$E = \frac{\sigma}{2\varepsilon_0} = \frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\text{ N/C}$$
The dipole moment is:
$$p = q \cdot d = (2 \times 10^{-6}\text{ C}) \times (0.2\text{ m}) = 4 \times 10^{-7}\text{ C}\cdot\text{m}$$
Substitute $p$, $E$, and $\theta$ into the torque formula:
$$\tau = pE\sin\theta$$
$$\tau = \left[(2 \times 10^{-6}) \times (0.2)\right] \times \left[\frac{100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}\right] \times \left(\frac{1}{2}\right)$$
$$\tau = \frac{10}{8.85} \approx 1.12\text{ N}\cdot\text{m}$$
{{SOL_IMG1}}
### Pattern Recognition
Equal and opposite charges on a rigid rod constitute an electric dipole. In a uniform field, the net translational force vanishes ($\vec{F}_{\text{net}} = \vec{0}$), leaving only a pure restoring torque $\tau = pE\sin\theta$.
### Evaluation Rubric / Model Answer
Option B: 1.12 Nm
### Chapter Mix
Class 12 Physics: Electrostatics
