Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 4.
Q14 (2025)
A capacitor, $C_1 = 6\mu\mathrm{F}$ is charged to a potential difference of $V_0 = 5\mathrm{V}$ using a $5\mathrm{V}$ battery. The battery is removed and another capacitor, $C_2 = 12\mu\mathrm{F}$ is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges $(q_1$ and $q_2)$ on the capacitors $C_1$ and $C_2$ when equilibrium condition is reached.
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- $q_{1} = 15\mu\mathrm{C}, q_{2} = 30\mu\mathrm{C}$
- $q_{1} = 30\mu\mathrm{C}, q_{2} = 15\mu\mathrm{C}$
- $q_{1} = 10\mu\mathrm{C}, q_{2} = 20\mu\mathrm{C}$
- $q_{1} = 20\mu\mathrm{C}, q_{2} = 10\mu\mathrm{C}$
### Related Formula
$$Q_{\text{total}} = C_1 V_0$$
$$V_c = \frac{Q_{\text{total}}}{C_1 + C_2}$$
$q = C \cdot V_c$
### Core Logic
1. **Initial Charge Calculation**:
Before closing the switch, capacitor $C_1$ accumulates total charge:
$$q_1' = C_1 \cdot V_0 = 6\mu\mathrm{F} \times 5\mathrm{V} = 30\mu\mathrm{C}$$
Uncharged capacitor $C_2$ holds $q_2' = 0$.
{{SOL_IMG1}}
2. **Redistribution at Equilibrium**:
Closing the switch causes parallel redistribution until they reach a common potential $V_c$:
{{SOL_IMG2}}
$$V_c = \frac{30\mu\mathrm{C} + 0}{6\mu\mathrm{F} + 12\mu\mathrm{F}} = \frac{30}{18} = \frac{5}{3}\mathrm{~V}$$
3. **Final Charges**:
$$q_1 = C_1 \cdot V_c = 6 \times \frac{5}{3} = 10\mu\mathrm{C}$$
$$q_2 = C_2 \cdot V_c = 12 \times \frac{5}{3} = 20\mu\mathrm{C}$$
Hence, the charges are $10\mu\mathrm{C}$ and $20\mu\mathrm{C}$ respectively.
### Pattern Recognition
For parallel combination setups, the final charge splits in the exact direct ratio of their capacitances: $q_1 : q_2 = C_1 : C_2 = 6 : 12 = 1 : 2$. Out of $30\mu\mathrm{C}$, the breakdown must cleanly be $10\mu\mathrm{C}$ and $20\mu\mathrm{C}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q1 (2025)
Two capacitors $C_1$ and $C_2$ are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are $U_1$ and $U_2$ , respectively. Which of the given statements is true?
{{IMG1}}
- $\mathrm{C}_1 > \mathrm{C}_2, \mathrm{U}_1 > \mathrm{U}_2$
- $\mathrm{C}_2 > \mathrm{C}_1, \mathrm{U}_2 < \mathrm{U}_1$
- $\mathrm{C}_1 > \mathrm{C}_2, \mathrm{U}_1 < \mathrm{U}_2$
- $\mathrm{C}_2 > \mathrm{C}_1, \mathrm{U}_2 > \mathrm{U}_1$
### Related Formula
$$\mathrm{V} = \text{same}$$
$$\mathrm{U} = \frac{1}{2} \mathrm{C}\mathrm{V}^2$$
$$\mathrm{q} = \mathrm{C}\mathrm{V}$$
### Core Logic
Since both capacitors are connected in parallel across the same battery, their potential difference $\mathrm{V}$ is identical.
From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first:
$$\mathrm{q}_2 > \mathrm{q}_1$$
Using the relation $\mathrm{q} = \mathrm{C}\mathrm{V}$, since $\mathrm{V}$ is identical, we get:
$$\mathrm{C}_2 > \mathrm{C}_1$$
Now, the electrostatic energy stored in a capacitor is given by:
$$\mathrm{U} = \frac{1}{2} \mathrm{C}\mathrm{V}^2$$
Since $\mathrm{C}_2 > \mathrm{C}_1$ and $\mathrm{V}$ is constant, the stored energy satisfies:
$$\mathrm{U}_2 > \mathrm{U}_1$$
### Step 1: Final Conclusion
Thus, the correct relationships are $\mathrm{C}_2 > \mathrm{C}_1$ and $\mathrm{U}_2 > \mathrm{U}_1$.
### Pattern Recognition
Sees parallel connection → Immediately lock potential difference $\mathrm{V}$ as constant. This simplifies $\mathrm{q} \propto \mathrm{C}$ and $\mathrm{U} \propto \mathrm{C}$, creating a direct linear bridge from graph height to energy capacity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q4 (2025)
Three infinitely long wires with linear charge density $\lambda$ are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?
- $\mathrm{xy} + \mathrm{yz} + \mathrm{zx} = \text{constant}$
- $(\mathrm{x + y})(\mathrm{y + z})(\mathrm{z + x}) = \text{constant}$
- $(\mathrm{x}^2 +\mathrm{y}^2)(\mathrm{y}^2 +\mathrm{z}^2)(\mathrm{z}^2 +\mathrm{x}^2) = \text{constant}$
- $\mathrm{xyz} = \text{constant}$
### Related Formula
$$\mathrm{V} = -\int \vec{\mathrm{E}}\cdot\mathrm{d}\vec{\mathrm{r}} = 2\mathrm{k}\lambda \ln \mathrm{r} + \mathrm{c}$$
### Core Logic
The potential at a point due to a line charge on an axis is proportional to the logarithm of its perpendicular distance.
For the wire along the z-axis: $\mathrm{V}_z = -\mathrm{k}\lambda \ln(x^2 + y^2)$
For the wire along the x-axis: $\mathrm{V}_x = -\mathrm{k}\lambda \ln(y^2 + z^2)$
For the wire along the y-axis: $\mathrm{V}_y = -\mathrm{k}\lambda \ln(z^2 + x^2)$
Summing the individual potentials to find the net configuration potential:
$$\mathrm{V}_{\text{net}} = -\mathrm{k}\lambda \left[ \ln(x^2+y^2) + \ln(y^2+z^2) + \ln(z^2+x^2) \right] + \mathrm{C}'$$
$$\mathrm{V}_{\text{net}} = -\mathrm{k}\lambda \ln \left[ (x^2+y^2)(y^2+z^2)(z^2+x^2) \right] + \mathrm{C}'$$
For an equipotential surface, set $\mathrm{V}_{\text{net}} = \text{constant}$:
### Step 1: Final Expression
$$(x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}$$
This maps perfectly to option (3).
### Pattern Recognition
Logarithmic combination rules transform scalar potential additions into products inside the functional argument: $\sum \ln(\mathrm{r}_i^2) = \ln(\prod \mathrm{r}_i^2)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q17 (2025)
A particle of mass $\mathrm{m}$ and charge $\mathrm{q}$ is fastened to one end $\mathrm{A}$ of a massless string having equilibrium length $\ell$ , whose other end is fixed at point $\mathrm{O}$ . The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x-axis is
{{IMG1}}
{{IMG2}}
- $\sqrt{\frac{2\mathrm{qE}\ell}{\mathrm{m}}}$
- $\sqrt{\frac{\mathrm{qE}\ell}{4\mathrm{m}}}$
- $\sqrt{\frac{q \mathrm{E} \ell}{m}}$
- $\sqrt{\frac{\mathrm{qE}\ell}{2\mathrm{m}}}$
### Core Logic
Applying the work-energy balance for the system as the particle moves from its initial position to the x-axis:
$$\mathrm{W}_{\text{all}} = \Delta \mathrm{k}$$
$$\mathrm{W}_{\mathrm{e}} = \mathrm{k}_{\mathrm{f}} - \mathrm{k}_{\mathrm{i}}$$
{{SOL_IMG1}}
The displacement parallel to the uniform electric field lines equals $\frac{\ell}{2}$. Therefore, computing the work done by the electrostatic field:
$$\mathrm{qE} \frac{\ell}{2} = \frac{1}{2} \mathrm{m v}^2 - 0$$
### Step 1: Final Expression
$$\mathrm{v} = \sqrt{\frac{\mathrm{qE}\ell}{\mathrm{m}}}$$
This matches option (3).
### Pattern Recognition
Keep your focus on displacement along the field lines. Work relies entirely on the parallel displacement component ($d_x = \ell \cos 60^{\circ}$), completely ignoring any vertical movement components.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electric Charges and Fields
Q758 (2025)
A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $\varepsilon_1$ and $\varepsilon_2$, as shown in figures. The distance between the plates is $d$ and area of each plate is $A$. If capacitance in first configuration and second configuration are $C_1$ and $C_2$ respectively, then $\frac{C_1}{C_2}$ is:
- $\frac{\varepsilon_1\varepsilon_2^2}{(\varepsilon_1 + \varepsilon_2)^2}$
- $\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$
- $\frac{\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2}$
- $\frac{\varepsilon_0(\varepsilon_1 + \varepsilon_2)}{2}$
### Related Formula
Capacitance with dielectric:
$$C = \frac{\varepsilon_r \varepsilon_0 A}{d}$$
Series Capacitors:
$$C_{\text{eq}} = \frac{C_a C_b}{C_a + C_b}$$
Parallel Capacitors:
$$C_{\text{eq}} = C_a + C_b$$
### Core Logic
Let $C_0 = \frac{\varepsilon_0 A}{d}$ be the capacitance without any dielectric.
- **First Configuration (Series connection)**:
The dielectrics split the gap vertically, so the effective thickness of each slab is $d/2$, and the area remains $A$.
$$C_a = \frac{\varepsilon_1 \varepsilon_0 A}{d/2} = 2\varepsilon_1 C_0$$
$$C_b = \frac{\varepsilon_2 \varepsilon_0 A}{d/2} = 2\varepsilon_2 C_0$$
Since they are in series:
$$C_1 = \frac{C_a C_b}{C_a + C_b} = \frac{(2\varepsilon_1 C_0)(2\varepsilon_2 C_0)}{2\varepsilon_1 C_0 + 2\varepsilon_2 C_0} = \frac{4\varepsilon_1\varepsilon_2 C_0^2}{2C_0(\varepsilon_1 + \varepsilon_2)} = \frac{2\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2} C_0$$
- **Second Configuration (Parallel connection)**:
The dielectrics split the area horizontally, so the effective area of each slab is $A/2$, and the distance remains $d$.
$$C_c = \frac{\varepsilon_1 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_1 C_0}{2}$$
$$C_d = \frac{\varepsilon_2 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_2 C_0}{2}$$
Since they are in parallel:
$$C_2 = C_c + C_d = (\varepsilon_1 + \varepsilon_2) \frac{C_0}{2}$$
{{SOL_IMG1}}
{{SOL_IMG2}}
### Step 1: Calculating the Ratio
Now, compute $\frac{C_1}{C_2}$:
$$\frac{C_1}{C_2} = \frac{\left(\frac{2\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2}\right) C_0}{\left(\frac{\varepsilon_1 + \varepsilon_2}{2}\right) C_0} = \frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$$
### Pattern Recognition
For dielectric-filled capacitors: splitting the gap ($d/2$) leads to a series combination, while splitting the plate area ($A/2$) leads to a parallel combination.
Shortcut: $C_{\text{series}} = \text{harmonic mean}$, $C_{\text{parallel}} = \text{arithmetic mean}$. The ratio $\frac{C_1}{C_2}$ is always the ratio of the harmonic mean of the dielectric constants to their arithmetic mean!
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
