Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 13.
Q42 (2026)
Electric field in a region is given by $\vec{E} = Ax\hat{i} + By\hat{j}$, where $A = 10~\mathrm{V / m^2}$ and $B = 5~\mathrm{V / m^2}$. If the electric potential at a point (10, 20) is $500~\mathrm{V}$, then the electric potential at origin is \_\_\_\_ V.
- 1000
- 500
- 2000
- 0
### Related Formula
$$V_2 - V_1 = -\int \vec{E} \cdot d\vec{r}$$
### Core Logic
Using potential difference relation:
$$500 - V_0 = -\int_{(0,0)}^{(10,20)} (10x\hat{i} + 5y\hat{j}) \cdot (dx\hat{i} + dy\hat{j})$$
$$500 - V_0 = -\left[5x^2 + \frac{5y^2}{2}\right]_{(0,0)}^{(10,20)}$$
$$V_0 - 500 = 500 + 1000 \implies V_0 = 2000 \text{ V}$$
### Pattern Recognition
Sees: Electric field vector function given, find potential at origin.
Shortcut: Integrate line integral of electric field from origin to given point.
Check: Matches option (3). ✓
### Chapter Mix
Class 12 Physics: Electrostatics
Q43 (2026)
A simple pendulum has a bob with mass m and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is \_\_\_\_.
(g : acceleration due to gravity)
- $mg - qE$
- $mg + qE$
- $\sqrt{m^2g^2 + q^2E^2}$
- $\sqrt{m^2g^2 - q^2E^2}$
### Related Formula
$$T = \sqrt{(qE)^2 + (mg)^2}$$
### Core Logic
{{SOL_IMG1}}
At equilibrium, the effective forces acting on the bob are vertical gravitational force $mg$ and horizontal electric force $qE$. The string tension balances the resultant of these orthogonal forces:
$$T = \sqrt{(qE)^2 + (mg)^2}$$
### Pattern Recognition
Sees: Charged pendulum in horizontal electric field.
Shortcut: Combine orthogonal forces ($mg$ downwards and $qE$ horizontally) via Pythagorean vector addition.
Check: Matches option (3). ✓
### Chapter Mix
Class 12 Physics: Electrostatics
Q30 (2026)
Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :
- $1:3^{1/3}$
- $1:2^{2/3}$
- $3^{2/3}:1$
- $1:3^{2/3}$
### Related Formula
$$V = \frac{kq}{r}$$
$$\text{Volume Conservation: } N \cdot \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$$
### Core Logic
From volume conservation of 3 coalescing droplets:
$$3 \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3 \implies R = 3^{1/3}r$$
Total charge on resultant bigger bubble $Q = 3q$.
Calculating initial potential $V_i$ and final potential $V_f$:
$$V_i = \frac{kq}{r}$$
$$V_f = \frac{k(3q)}{R} = \frac{3kq}{3^{1/3}r} = 3^{2/3} \frac{kq}{r}$$
Ratio of initial to final potential:
$$\frac{V_i}{V_f} = \frac{1}{3^{2/3}} = 1 : 3^{2/3}$$
{{SOL_IMG1}}
### Step 1: Final Conclusion
The ratio of potentials is $1 : 3^{2/3}$.
### Pattern Recognition
Coalescing droplets rule: For $N$ identical drops, $R = N^{1/3}r$ and $Q = Nq$.
Potential ratio $V_i / V_f = 1 / N^{2/3}$. For $N=3$, ratio is $1 / 3^{2/3}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q42 (2026)
Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure. The electric potential (V) and the electric field ($\vec{E}$) at the center O of the pentagon due to these five positive charges are :
{{IMG1}}
- $V = \frac{5q}{4\pi\varepsilon_0r}$ and $\vec{E} = 0$
- $V = \frac{5q}{4\pi\varepsilon_{0}r}$ and $\vec{E} = \frac{5\sqrt{3}q}{8\pi\varepsilon_{0}r^{2}} \hat{r}$
- $V = \frac{5q}{4\pi\varepsilon_0r}$ and $\vec{E} = \frac{5q}{4\pi\varepsilon_0r^2}\hat{r}$
- $V = 0$ and $\vec{E} = 0$
### Related Formula
$$V = \sum \frac{k q_i}{r}$$
$$\vec{E}_{\text{center}} = \sum \vec{E}_i = 0 \quad (\text{Symmetric Polygon})$$
### Core Logic
Due to spatial symmetry of identical charges at the 5 vertices of a regular pentagon, vector sum of electric fields at center O cancels out:
$\vec{E} = 0$
Electric potential is a scalar sum:
$$V = 5 \times \left(\frac{q}{4\pi\varepsilon_0 r}\right) = \frac{5q}{4\pi\varepsilon_0 r}$$
### Step 1: Final Conclusion
Option (1) gives the correct values $V = \frac{5q}{4\pi\varepsilon_0r}$ and $\vec{E} = 0$.
### Pattern Recognition
Symmetry rule: Identical charges at vertices of any regular polygon $\implies \vec{E}_{center} = 0$. Potential is scalar addition $V = N \frac{kq}{r}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q48 (2026)
A capacitor P with capacitance $10 \times 10^{-6}$ F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance $20 \times 10^{-6}$ F. The charge on capacitor Q when equilibrium is established will be $\alpha \times 10^{-5}$ C (assume capacitor Q does not have any charge initially), the value of $\alpha$ is ____.
### Related Formula
$$V_{\text{common}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$
$$Q_2 = C_2 V_{\text{common}}$$
### Core Logic
Given $C_1 = 10 \times 10^{-6} \mathrm{~F}, V_1 = 6.0 \mathrm{~V}$ and $C_2 = 20 \times 10^{-6} \mathrm{~F}, V_2 = 0 \mathrm{~V}$:
$$V_{\text{common}} = \frac{10^{-5} \times 6 + 0}{10^{-5} + 2 \times 10^{-5}} = \frac{6 \times 10^{-5}}{3 \times 10^{-5}} = 2 \mathrm{~V}$$
Calculating final charge on capacitor Q ($C_2$):
$$Q_2 = C_2 V_{\text{common}} = (20 \times 10^{-6} \mathrm{~F}) \times 2 \mathrm{~V} = 40 \times 10^{-6} \mathrm{~C} = 4 \times 10^{-5} \mathrm{~C}$$
Comparing with $\alpha \times 10^{-5} \mathrm{~C} \implies \alpha = 4$.
### Step 1: Final Conclusion
The value of $\alpha$ is 4.
### Pattern Recognition
Charge distribution rule: Total initial charge $Q_{total} = C_1 V_1 = 60\,\mu\mathrm{C}$.
Final charge splits in proportion to capacitance ratio $C_2 / (C_1+C_2) = 2/3$.
$Q_2 = (2/3) \times 60\,\mu\mathrm{C} = 40\,\mu\mathrm{C} = 4 \times 10^{-5}\mathrm{~C}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
