Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 8.
Q2 (2025)
A parallel plate capacitor of capacitance $1\mu \mathrm{F}$ is charged to a potential difference of $20\mathrm{V}$. The distance between plates is $1\mu \mathrm{m}$. The energy density between plates of capacitor is:
- $1.8 \times 10^{3} \, \mathrm{J/m}^{3}$
- $2 \times 10^{-4} \, \mathrm{J/m^3}$
- $2 \times 10^{2} \mathrm{~J} / \mathrm{m}^{3}$
- $1.8 \times 10^{5} \mathrm{~J} / \mathrm{m}^{3}$
### Related Formula
The electric field $E$ between the plates of a parallel plate capacitor is given by :
$$E = \frac{V}{d}$$
The electrostatic energy density $U$ in a medium is given by:
$$U = \frac{1}{2} \epsilon_0 E^2$$
### Core Logic
Given values [cite: 662, 664, 665]:
* Capacitance, $C = 1 \ \mu\text{F}$
* Potential difference, $V = 20 \text{ V}$
* Plate separation, $d = 1 \ \mu\text{m} = 10^{-6} \text{ m}$
First, calculate the electric field $E$ :
$$E = \frac{20}{10^{-6}} = 20 \times 10^6 \text{ V/m}$$
Now, calculate the energy density $U$ using $\epsilon_0 \approx 8.85 \times 10^{-12} \text{ F/m}$ :
$$U = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (20 \times 10^6)^2$$
$$U = \frac{1}{2} \times 8.85 \times 10^{-12} \times 400 \times 10^{12}$$
$$U = 8.85 \times 200 = 1770 \text{ J/m}^3 = 1.77 \times 10^3 \text{ J/m}^3$$
Rounding to the nearest matching option gives $1.8 \times 10^3 \text{ J/m}^3$.
### Pattern Recognition
Notice that energy density is completely independent of the total capacitance value $C$ if the voltage and spacing are directly provided. Always check for redundant parameters included to distract candidates.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q22 (2025)
An electric dipole of dipole moment $6 \times 10^{-6} \mathrm{Cm}$ is placed in uniform electric field of magnitude $10^{6} \mathrm{~V/m}$. Initially, the dipole moment is \parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ______ J.
### Related Formula
The potential energy of an electric dipole aligned at an \angle $\theta$ inside a uniform electric field is given by:
$$U = -p E \cos \theta$$
The work done by an external agent to rotate the dipole equals its change in potential energy:
$$W = \Delta U = -pE (\cos \theta_f - \cos \theta_i)$$
### Core Logic
Given parameters [cite: 805, 806]:
* Dipole moment, $p = 6 \times 10^{-6} \text{ Cm}$
* Electric field, $E = 10^6 \text{ V/m}$
* Initial \angle (\parallel state), $\theta_i = 0^{\circ}$
* Final \angle (opposite state), $\theta_f = 180^{\circ}$
Substitute the values into the work equation [cite: 807, 808]:
$$W = -pE (\cos 180^{\circ} - \cos 0^{\circ})$$
$$W = -pE (-1 - 1) = 2 p E \quad \text{}$$
$$W = 2 \times (6 \times 10^{-6}) \times 10^6 = 12 \text{ J} \quad \text{}$$
### Pattern Recognition
Rotating a dipole from its most stable configuration (\parallel, $\theta=0^{\circ}$) to its most unstable position (anti-\parallel, $\theta=180^{\circ}$) always requires a maximum work value of exactly $2pE$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q1278 (2025)
An electric dipole of mass m, charge q, and length $l$ is placed in a uniform electric field $\vec{\mathrm{E}} = \mathrm{E}_0\hat{\mathrm{i}}$ . When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:
- $\frac{1}{2\pi}\sqrt{\frac{2\mathrm{ml}}{\mathrm{qE}_0}}$
- $2\pi \sqrt{\frac{\mathrm{m}l}{\mathrm{qE}_0}}$
- $\frac{1}{2\pi} \sqrt{\frac{m l}{2 q E_{0}}}$
- $2\pi \sqrt{\frac{\mathrm{m}l}{2\mathrm{qE}_0}}$
### Related Formula
$$\tau = -pE \sin \theta$$
$$I = 2 m \left(\frac{l}{2}\right)^2 = \frac{ml^2}{2}$$
### Core Logic
Restoring torque for small angle $\theta$ is given by:
$$\tau = -q l E_0 \theta$$
Equating with rotational inertia dynamics:
$$I \omega^2 \theta = q l E_0 \theta$$
$$\frac{m l^2}{2} \omega^2 = q l E_0 \implies \omega^2 = \frac{2 q E_0}{m l}$$
### Step 1: Compute Time Period
$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{ml}{2qE_0}}$$
### Pattern Recognition
For a two-particle system pivoting about midpoint, total moment of inertia drops to $ml^2/2$, scaling the period by a factor of $\sqrt{2}$
### Chapter Mix
Class 12 Physics: Electrostatics
Q1293 (2025)
Match List-I with List-II.
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I</th><th style="border: 1px solid #888; padding: 8px;">List-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">(A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.</td><td style="border: 1px solid #888; padding: 8px;">(I) $\sigma / \epsilon_0$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(B) Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σ.</td><td style="border: 1px solid #888; padding: 8px;">(II) $\sigma / 2\epsilon_0$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(C) Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R</td><td style="border: 1px solid #888; padding: 8px;">(III) 0</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ.</td><td style="border: 1px solid #888; padding: 8px;">(IV) $\sigma R^2 / \epsilon_0 r^2$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
- (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
- (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
- (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
### Core Logic
Mapping electrostatics equations via Gauss\'s law applications :
* (A) Inside a shell, enclosed charge is zero $\implies E = 0$ (III) .
* (B) Near an infinite sheet, $E = \frac{\sigma}{2\epsilon_0}$ (II) .
* (C) Outside a shell, $E = \frac{kQ}{r^2} = \frac{\sigma R^2}{\epsilon_0 r^2}$ (IV) .
* (D) Between opposite sheets, fields add up: $\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$ (I) .
Hence, the proper combination sequence is (A)-(III), (B)-(II), (C)-(IV), (D)-(I).
### Chapter Mix
Class 12 Physics: Electrostatics
Q1374 (2024)
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5\mathrm{~g/cc}$, the dielectric constant of water will be (Take density of water $= 1\mathrm{~g/cc}$):
### Related Formula
Equilibrium condition for electrostatic suspension:
$$\tan\left(\frac{\theta}{2}\right) = \frac{F_e}{mg} = \frac{q^2}{4\pi\varepsilon_0 r^2 mg}$$
In a liquid medium with buoyant force mitigation:
$$\tan\left(\frac{\theta}{2}\right) = \frac{F_e'}{mg_{\text{eff}}} = \frac{q^2}{4\pi\varepsilon_0 \varepsilon_r r^2 mg \left(1 - \frac{\rho_{\text{liquid}}}{\rho_{\text{solid}}}\right)}$$
### Core Logic
Since the angle $\theta$ stays exactly the same in both scenarios, we can equate the two balance ratios:
$$\frac{F_e}{mg} = \frac{F_e'}{mg_{\text{eff}}} \implies 1 = \varepsilon_r \left(1 - \frac{\rho_w}{\rho_s}\right)$$
### Step 1: Substitute Densities
Given data: $\rho_s = 1.5\mathrm{~g/cc}$, $\rho_w = 1.0\mathrm{~g/cc}$.
$$1 = \varepsilon_r \left(1 - \frac{1}{1.5}\right) = \varepsilon_r \left(1 - \frac{2}{3}\right) = \varepsilon_r \left(\frac{1}{3}\right)$$
$$\varepsilon_r = 3$$
### Pattern Recognition
Shortcut formula for invariant angle setups:
$$\varepsilon_r = \frac{\rho_{\text{solid}}}{\rho_{\text{solid}} - \rho_{\text{liquid}}} = \frac{1.5}{1.5 - 1} = \frac{1.5}{0.5} = 3$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Class 11 Physics: Mechanical Properties of Fluids
