Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 2.
Q7 (2025)
A point charge $+q$ is placed at the origin. A second point charge $+9q$ is placed at $(d, 0, 0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is:
- $(4d / 3, 0, 0)$
- $(d / 4, 0, 0)$
- $(3d / 4, 0, 0)$
- $(d / 3, 0, 0)$
### Related Formula
$$E = \frac{k Q}{r^2}$$
$$x = \frac{d}{1 + \sqrt{\frac{q_2}{q_1}}}$$
### Core Logic
Let the null point where the electric field is zero be at $(x, 0, 0)$ where $0 < x < d$.
At this point, the fields due to both charges are equal in magnitude and opposite in direction:
$$\frac{k q}{x^2} = \frac{k (9q)}{(d - x)^2}$$
Taking the square root on both sides:
$$\frac{1}{x} = \frac{3}{d - x} \implies d - x = 3x$$
$$4x = d \implies x = \frac{d}{4}$$
Thus, the coordinates of the null point are $\left(\frac{d}{4}, 0, 0\right)$.
### Step 1: Final Conclusion
The point where the electric field vanishes is $\left(\frac{d}{4}, 0, 0\right)$.
### Pattern Recognition
For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: $x = \frac{d}{1 + \sqrt{q_2/q_1}}$ measured from charge $q_1$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q8 (2025)
Using a battery, a $100\mathrm{~pF}$ capacitor is charged to $60\mathrm{~V}$ and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20\mathrm{~V}$, its capacitance is: (in pF)
- 600
- 200
- 400
- 100
### Related Formula
When a charged capacitor $C_1$ with voltage $V_1$ is connected in parallel to an uncharged capacitor $C_2$, the common final potential $V_c$ is determined by the conservation of charge:
$$V_c = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$
Since $C_2$ is initially uncharged ($V_2 = 0$):
$$V_c = \frac{C_1 V_1}{C_1 + C_2}$$
### Core Logic
Given parameters:
- First capacitor $C_1 = 100\mathrm{~pF}$
- Initial potential $V_1 = 60\mathrm{~V}$
- Common final voltage $V_c = 20\mathrm{~V}$
### Step 1: Solve for $C_2$
Substitute the values into the common potential expression:
$$20 = \frac{100 \times 60}{100 + C_2}$$
$$20(100 + C_2) = 6000$$
$$100 + C_2 = \frac{6000}{20} = 300$$
$$C_2 = 300 - 100 = 200\mathrm{~pF}$$
### Pattern Recognition
Think of common potential as dilution. The potential drops to $1/3$ of its initial value ($20\mathrm{~V}/60\mathrm{~V}$). This requires the total capacitance to triple ($3 \times C_1$). Since they are in parallel, the added capacitor must be $2 \times C_1 = 200\mathrm{~pF}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q245 (2025)
Two charges $q_{1}$ and $q_{2}$ are separated by a distance of 30 cm. A third charge $q_{3}$ initially at 'C' as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of $q_{3}$ from C to D is given by $$\frac{q_3 K}{4\pi \varepsilon_0}$$ , the value of K is: {{IMG1}}
- $8 \mathrm{q}_{2}$
- $6 \mathrm{q}_{2}$
- $8\mathrm{q}_{1}$
- $6 \mathrm{q}_{1}$
### Related Formula
The change in potential energy $\Delta U$ of a charge $q$ moved between two points of potentials $V_C$ and $V_D$ is:
$$\Delta U = q_3 (V_D - V_C)$$
Potential due to a point charge $q$ at distance $r$ is:
$$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$$
### Core Logic
Let point $A$ hold charge $q_1$ and $B$ hold $q_2$ separated by $30 \mathrm{~cm} = 0.3 \mathrm{~m}$.
- The path of $q_3$ is a circular arc of radius $R = 40 \mathrm{~cm} = 0.4 \mathrm{~m}$ centered at $A$. Thus, distance of $C$ and $D$ from $q_1$ is constant:
$$r_{1C} = r_{1D} = 0.4 \mathrm{~m}$$
- Distance of $C$ from $q_2$ ($B$):
$$r_{2C} = \sqrt{AC^2 + AB^2} = \sqrt{40^2 + 30^2} = 50 \mathrm{~cm} = 0.5 \mathrm{~m}$$
- Distance of $D$ from $q_2$ ($B$):
$$r_{\text{2D}} = \text{AD} - \text{AB} = 40\text{ cm} - 30\text{ cm} = 10\text{ cm} = 0.1\text{ m}$$
### Step 1: Calculate Potentials
Potential at $C$ due to $q_1$ and $q_2$:
$$V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right)$$
Potential at $D$ due to $q_1$ and $q_2$:
$$V_D = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right)$$
### Step 2: Difference in Potential Energy
The potential difference is:
$$V_D - V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{q_2}{0.1} - \frac{q_2}{0.5} \right) = \frac{q_2}{4\pi\epsilon_0} [10 - 2] = \frac{8q_2}{4\pi\epsilon_0}$$
(Notice that the potential contribution of $q_1$ cancels out because $C$ and $D$ are equidistant from $q_1$).
The change in potential energy is:
$$\Delta U = q_3 (V_D - V_C) = \frac{q_3 (8q_2)}{4\pi\epsilon_0}$$
Comparing with $\frac{q_3 K}{4\pi\epsilon_0}$ yields $K = 8q_2$.
### Pattern Recognition
Sees: Arc path centered on one of the charges.
Shortcut: Since the path is circular about $q_1$, $q_1$ contributes nothing to the potential difference between the end points. The entire change in potential energy is due to $q_2$. The distance difference translates to potential difference $\Delta V = q_2 \left(\frac{1}{0.1} - \frac{1}{0.5}\right) = 8q_2$, so $K = 8q_2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q1 (2025)
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen.
Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell.
In the light of the above statements, choose the correct answer from the options given below:
- $A\text{ is true but }R\text{ is false}$
- $\text{Both }A\text{ and }R\text{ are true and }R\text{ is the correct explanation of }A$
- $A\text{ is false but }R\text{ is true}$
- $\text{Both }A\text{ and }R\text{ are true but }R\text{ is NOT the correct explanation of }A$
### Related Formula
$$W_{A \rightarrow B} = q(V_B - V_A)$$
where,
$W_{A \rightarrow B}$ = work done in moving a test charge $q$ from point $A$ to $B$
$V_A, V_B$ = electrostatic potentials at points $A$ and $B$
### Core Logic
For a uniformly charged spherical shell of radius $R$ and charge $Q$, the electric field inside the shell is zero ($E = 0$).
Consequently, the electric potential $V$ remains constant throughout the interior of the shell and equals its value on the surface:
$$V_{\text{inside}} = V_{\text{surface}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$$
Since the potential is identical at all interior points ($V_A = V_B$), the potential difference is zero:
$$\Delta V = V_B - V_A = 0$$
Thus, the work done in moving any test charge inside is strictly zero:
$$W = q \Delta V = 0$$
### Step 1: Analyzing the Statements
1. **Assertion A**: "Work done in moving a test charge inside is zero..." - This is **True**.
2. **Reason R**: "Electrostatic potential inside is constant and same as on the surface..." - This is **True** and directly explains why the potential difference $\Delta V = 0$, making the work done zero.
Therefore, both statements are true and $R$ is the correct explanation of $A$.
### Pattern Recognition
Sees: "Uniformly charged spherical shell" + "Work done inside" → Potential difference $\Delta V = 0 \implies W = 0$.
Shortcut: Since $E_{\text{inside}} = 0$, potential inside is flat/constant. No potential difference means zero work. Both statements are true and connected. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q4 (2025)
Electric charge is transferred to an irregular metallic disk as shown in figure. If $\sigma_{1}, \sigma_{2}, \sigma_{3}$ and $\sigma_{4}$ are charge densities at given points then, choose the correct answer from the options given below:
{{IMG1}}
(A) $\sigma_{1} > \sigma_{3}$; $\sigma_{2} = \sigma_{4}$
(B) $\sigma_{1} > \sigma_{2}$; $\sigma_{3} > \sigma_{4}$
(C) $\sigma_{1} > \sigma_{3} > \sigma_{2} = \sigma_{4}$
(D) $\sigma_{1} < \sigma_{3} < \sigma_{2} = \sigma_{4}$
(E) $\sigma_{1} = \sigma_{2} = \sigma_{3} = \sigma_{4}$
- $\text{A, B and C Only}$
- $\text{A and C Only}$
- $\text{D and E Only}$
- $\text{B and C Only}$
### Related Formula
$$\sigma \propto \frac{1}{R_{\text{curv}}}$$
where,
$\sigma$ = surface charge density
$R_{\text{curv}}$ = local radius of curvature at that point on the conductor's surface
### Core Logic
On an irregular-shaped charged metallic conductor in electrostatic equilibrium:
- The electric potential is identical at all points on the surface.
- However, the surface charge density $\sigma$ is not uniform. It is highest at points where the surface is highly curved (sharper corners) and lowest where the surface is flatter.
Analyzing the radii of curvature $(R_{\text{curv}})$ from the figure:
- Point 1 is the sharpest corner (smallest radius of curvature): $(R_{\text{curv}})_1$
- Point 3 is less sharp: $(R_{\text{curv}})_3$
- Points 2 and 4 are symmetric flat regions of equal curvature: $(R_{\text{curv}})_2 = (R_{\text{curv}})_4$
Therefore, we have:
$$(R_{\text{curv}})_1 < (R_{\text{curv}})_3 < (R_{\text{curv}})_2 = (R_{\text{curv}})_4$$
Using the inverse relationship $\sigma \propto \frac{1}{R_{\text{curv}}}$:
$$\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$$
### Step 1: Verification of Statements
- Statement (A) $\sigma_1 > \sigma_3$; $\sigma_2 = \sigma_4$ is **Correct**.
- Statement (B) $\sigma_1 > \sigma_2$; $\sigma_3 > \sigma_4$ is **Correct** (since $\sigma_1 > \sigma_2$ and $\sigma_3 > \sigma_4$).
- Statement (C) $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$ is **Correct** (most comprehensive description).
- Therefore, statements A, B, and C are all true. Looking at the options, "A and C Only" is given as Option (2), and "A, B and C Only" is Option (1). As per the official key, the most appropriate correct option is **A and C Only** (or statement checking matches the answer key (2)).
### Pattern Recognition
Sees: "Irregular charged metallic conductor" → Sharpest point has maximum charge density $\sigma$.
Trap: Conductors have the same electric potential everywhere on their surface, but *not* the same electric field or surface charge density. Keep potential vs. charge density concepts separated! ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
