Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 7.
Q7 (2025)
In the first configuration (1) as shown in the figure {{IMG1}}, four identical charges $(q_{0})$ are kept at the corners A, B, C and D of square of side length 'a'.
In the second configuration (2) {{IMG2}}, the same charges are shifted to mid points G, E, H and F, of the square. If $K=\frac{1}{4\pi\varepsilon_{0}}$, the difference between the potential energies of configuration (2) and (1) is given by:
- $\frac{Kq_{0}^{2}}{a}\left(4\sqrt{2}-2\right)$
- $\frac{Kq_{0}^{2}}{a}\left(3-\sqrt{2}\right)$
- $\frac{Kq_{0}^{2}}{a}\left(4-2\sqrt{2}\right)$
- $\frac{Kq_{0}^{2}}{a}\left(3\sqrt{2}-2\right)$
### Related Formula
$$U = \sum_{i < j} \frac{K q_i q_j}{r_{ij}}$$
### Core Logic
For configuration (1) with side length $a$:
- 4 pairs of adjacent side-charges with distance $a$.
- 2 pairs of diagonal charges with distance $\sqrt{2}a$.
$$U_1 = 4 \cdot \frac{Kq_0^2}{a} + 2 \cdot \frac{Kq_0^2}{\sqrt{2}a} = \frac{Kq_0^2}{a}(4 + \sqrt{2})$$
For configuration (2), charges lie on midpoints forming an inner square of side length $a' = \frac{a}{\sqrt{2}}$:
- 4 pairs at distance $\frac{a}{\sqrt{2}}$.
- 2 pairs at diagonal distance $a$.
$$U_2 = 4 \cdot \frac{Kq_0^2}{\frac{a}{\sqrt{2}}} + 2 \cdot \frac{Kq_0^2}{a} = \frac{Kq_0^2}{a}(4\sqrt{2} + 2)$$
### Step 1: Finding the difference
$$U_2 - U_1 = \frac{Kq_0^2}{a}\left[(4\sqrt{2} + 2) - (4 + \sqrt{2})\right] = \frac{Kq_0^2}{a}(3\sqrt{2} - 2)$$
### Pattern Recognition
The side of the new square formed by midpoints is scaled down by $1/\sqrt{2}$. Therefore, interaction terms scale accordingly based on system geometric dimensions.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q18 (2025)
A small uncharged conducting sphere is placed in contact with an identical sphere but having $4 \times 10^{-8}$ C charge and then removed to a distance such that the force of repulsion between them is $9 \times 10^{-3}$ N. The distance between them is (Take $\frac{1}{4\pi\varepsilon_{0}}$ as $9 \times 10^{9}$ in SI units)
- 2 cm
- 3 cm
- 4 cm
- 1 cm
### Related Formula
$$F = \frac{k q_1 q_2}{r^2}$$
### Core Logic
When two identical conducting spheres are brought into contact, the total initial charge splits equally between them:
$$q_1 = q_2 = \frac{4 \times 10^{-8}\ \mathrm{C} + 0}{2} = 2 \times 10^{-8}\ \mathrm{C}$$
Given repulsion force, $F = 9 \times 10^{-3}\ \mathrm{N}$:
$$9 \times 10^{-3} = \frac{9 \times 10^{9} \times (2 \times 10^{-8}) \times (2 \times 10^{-8})}{r^2}$$
$$9 \times 10^{-3} = \frac{36 \times 10^{-7}}{r^2} \implies r^2 = \frac{36 \times 10^{-7}}{9 \times 10^{-3}} = 4 \times 10^{-4}$$
$$r = 2 \times 10^{-2}\ \mathrm{m} = 2\ \mathrm{cm}$${{SOL_IMG1}}
### Pattern Recognition
Identical spheres in contact distribute net charge equally due to symmetric capacitance sharing: $q' = Q_{\text{total}} / 2$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatics
Q1126 (2025)
Consider a parallel plate capacitor of area A (of each plate) and separation 'd' between the plates. If E is the electric field and $\epsilon_{0}$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is :-
- $\frac{1}{2}\epsilon_{0}E^{2}Ad$
- $\frac{3}{4}\epsilon_{0}E^{2}Ad$
- $\frac{1}{4}\epsilon_{0}E^{2}Ad$
- $\epsilon_{0}E^{2}Ad$
### Related Formula
The electrostatic energy density $u$ stored in an electric field $E$ is given by:
$$u = \frac{1}{2}\epsilon_{0}E^{2}$$
The total potential energy $U$ stored in a volume $V$ is:
$U = u \cdot V$
### Core Logic
For a parallel plate capacitor, the volume between the plates where the electric field exists is the product of the plate area $A$ and the plate separation $d$:
$V = Ad$
### Step 1: Calculating Stored Energy
Substitute the volume expression into the total energy equation:
$$U = \left(\frac{1}{2}\epsilon_{0}E^{2}\right)(Ad)$$
$$U = \frac{1}{2}\epsilon_{0}E^{2}Ad$$
### Pattern Recognition
Energy density times volume is a universal relation for fields. Remember that volume is simply cross-sectional area multiplied by distance ($Ad$).
Q1146 (2025)
A square loop of sides $a = 1$ m is held normally in front of a point charge $q = 1C$ The flux of the electric field through the shaded region is $\frac{5}{p} \times \frac{1}{\varepsilon_0} \frac{Nm^2}{C}$ , where the value of p is . {{IMG1}}
### Related Formula
By Gauss's Law, the total flux emitted by a point charge $q$ through a completely enclosing symmetric cube container surface is:
$$\Phi_{\text{total}} = \frac{q}{\epsilon_{0}}$$
### Core Logic
Assuming the charge resides at a symmetric center distance $\frac{a}{2}$ relative to the loop face, this square loop represents one of the six identical faces of an enclosing cube system. Thus, the flux passing through the entire square loop face is:
$$\Phi_{\text{square}} = \frac{1}{6} \Phi_{\text{total}} = \frac{q}{6\epsilon_{0}}$$
### Step 1: Symmetric Partitioning
As shown in the solution schematic {{SOL_IMG1}}, the square face can be divided into 8 identical symmetric right-angled triangle sections by drawing its diagonals and medians. Each individual part intercepts an equal portion of the flux field :
$$\Phi_{\text{part}} = \frac{1}{8} \Phi_{\text{square}} = \frac{1}{8} \left(\frac{q}{6\epsilon_{0}}\right) = \frac{q}{48\epsilon_{0}}$$
The shaded region covers exactly 5 of these individual triangle parts :
$$\Phi_{\text{shaded}} = 5 \times \Phi_{\text{part}} = \frac{5}{48} \times \frac{q}{\epsilon_{0}}$$
Comparing this result with the given expression $\frac{5}{p} \times \frac{1}{\epsilon_{0}}$ , we find:
$p = 48$
### Pattern Recognition
Exploit geometric symmetry to break solid angles down into equal fractions, avoiding complex surface integration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q18 (2025)
A parallel plate capacitor was made with two rectangular plates, each with a length of $l = 3$ cm and breath of $b = 1$ cm. The distance between the plates is $3\mu m$ Out of the following, which are the ways to increase the capacitance by a factor of 10?
A. $l = 30$ cm, b = 1 cm, $d=1\mu m$
B. $l = 3$ cm, $b=1$ cm, $d=30~\mu m$
C. $l = 6$ cm, $b=5$ cm, $d=3~\mu m$
D. $l = 1$ cm, $b=1\text{cm}$, $d=10~\mu m$
E. $l = 5\text{ cm}$, $b=2$ cm, $d=1\mu m$
Choose the correct answer from the options given below :
- C and E only
- B and D only
- A only
- C only
### Related Formula
The capacitance of a parallel plate system is given by :
$$C = \frac{\epsilon_{0}A}{d} = \frac{\epsilon_{0}lb}{d}$$
where $l$ is length, $b$ is breadth, and $d$ is separation distance.
### Core Logic
Evaluate the initial capacitance base scaling parameter :
$$C_{0} = \frac{\epsilon_{0} \times 3\text{ cm} \times 1\text{ cm}}{3\mu\text{m}} = 1 \times \epsilon_{0}\text{ units}$$
We want to increase this initial baseline capacitance value by a factor of 10, meaning our target capacitance is $10\epsilon_{0}$.
### Step 1: Audit Options
Let's check the capacitance for options C and E :
* Option C: $l=6\text{ cm}, b=5\text{ cm}, d=3\mu\text{m}$ .
$$C_{C} = \frac{\epsilon_{0} \times 6 \times 5}{3} = 10\epsilon_{0}\text{ units (Correct)}$$
* Option E: $l=5\text{ cm}, b=2\text{ cm}, d=1\mu\text{m}$ .
$$C_{E} = \frac{\epsilon_{0} \times 5 \times 2}{1} = 10\epsilon_{0}\text{ units (Correct)}$$
### Pattern Recognition
Capacitance scales matching the geometric factor $\frac{l \cdot b}{d}$. Look for options where this ratio scales up to exactly 10 times the initial baseline value.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
