Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 6.
Q924 (2025)
Four capacitor each of capacitance $16mumathrm{F}$ are connected as shown in the figure. The capacitance between points A and B is: (in $mumathrm{F}$).
{{IMG1}}
### Related Formula
Equivalent capacitance for capacitors in parallel:
$$C_{\text{eq}} = C_1 + C_2 + C_3 + \dots$$
### Core Logic
By tracing electric potential along the ideal connecting wires, label the two plates of each capacitor with their connected terminal ($A$ or $B$).
{{SOL_IMG1}}
Redrawing the circuit reveals that all 4 capacitors are connected directly in parallel across terminals $A$ and $B$.
### Step 1: Calculate Equivalent Value
Since all four identical capacitors are connected in parallel:
$$C_{\text{eq}} = 4C$$
Given each capacitor has $C = 16\,\mu\text{F}$:
$$C_{\text{eq}} = 4 \times 16 = 64\,\mu\text{F}$$
{{SOL_IMG2}}
### Pattern Recognition
Bridging wires across alternate terminals typically simplifies an apparent ladder into a pure parallel arrangement. Always apply the node-potential labeling method first to identify terminals sharing identical potentials.
### Evaluation Rubric / Model Answer
64
### Chapter Mix
Class 12 Physics: Electrostatics
Q1 (2025)
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The outer body of an air craft is made of metal which protects persons sitting inside from lightning-strikes.
Reason (R): The electric field inside the cavity enclosed by a conductor is zero.
In the light of the above statements, chose the most appropriate answer from the options given below:
- Both (A) and (R) are correct and (R) is the correct explanation of (A)
- (A) is correct but (R) is not correct
- Both (A) and (R) are correct but (R) is not correct explanation of (A)
- (A) is not correct but (R) is correct
### Core Logic
According to electrostatic shielding, the electric field inside a cavity of a conductor is always zero, regardless of the size and shape of the cavity and regardless of any charges located outside or on the conductor's surface. Therefore, when lightning strikes a metal aircraft, the entire charge stays on the outer metallic surface and flows down without producing an electric field inside, keeping passengers safe.
### Step 1: Statement Evaluation
* **Assertion (A):** Correct, passengers are protected from lightning because of the metallic body shield.
* **Reason (R):** Correct, the field inside a cavity of a conductor is zero.
* **Explanation:** Since the zero field property is precisely why passengers are protected, (R) correctly explains (A).
### Pattern Recognition
Metallic shell / shield configuration always establishes $E_{\text{inside}} = 0$. This shielding mechanism directly underpins safety features in lightning scenarios for cars and airplanes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q10 (2025)
A dipole with two electric charges of 2 µC magnitude each, with separation distance 0.5 µm, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of 5 V is applied.
Separation between the plates is 0.5 mm. If the dipole is rotated by $30^{circ}$ from the axis, it tends to realign in the direction due to a torque.
The value of torque is : [cite: 58, 59, 60]
- $5\times10^{-9}~\mathrm{Nm}$ [cite: 60]
- $2.5\times10^{-12}~\mathrm{Nm}$ [cite: 61]
- $5\times10^{-3}~\mathrm{Nm}$ [cite: 62]
- $2.5\times10^{-9}~\mathrm{Nm}$ [cite: 62]
### Related Formula
$$E = \frac{V}{d}$$ [cite: 736]
$$\tau = pE\sin\theta$$ [cite: 737]
$p = q \cdot a$ [cite: 738]
### Core Logic
First, calculate the electric field magnitude $E$ between the capacitor plates: [cite: 58, 736]
$$E = \frac{5}{0.5 \times 10^{-3}} = 10^4\ \text{V/m}$$ [cite: 58, 59, 736]
Next, evaluate the dipole moment $p$: [cite: 58, 738]
$$p = (2 \times 10^{-6}\ \text{C}) \times (0.5 \times 10^{-6}\ \text{m}) = 1 \times 10^{-12}\ \text{C}\cdot\text{m}$$ [cite: 58, 740]
Now find the torque when rotated by $\theta = 30^{\circ}$: [cite: 59, 737]
$$\tau = (1 \times 10^{-12}) \times 10^4 \times \sin 30^{\circ} = 10^{-8} \times \frac{1}{2} = 5 \times 10^{-9}\ \text{N}\cdot\text{m}$$ [cite: 743]
### Pattern Recognition
Always convert parameters to pristine standard SI units before applying electrostatic expressions ($0.5\ \mu\text{m} = 5 \times 10^{-7}\ \text{m}$ and $0.5\ \text{mm} = 5 \times 10^{-4}\ \text{m}$) to secure zero conversion error[cite: 58, 59, 736, 738].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q21 (2025)
A parallel plate capacitor has charge $5 imes10^{-6}\mathrm{~C}$. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 imes10^{-6}\mathrm{~C}$ then the dielectric constant of the slab is _______. [cite: 183, 184]
### Related Formula
$$Q_{\text{ind}} = Q\left(1 - \frac{1}{K}\right)$$ [cite: 813]
### Core Logic
Substitute the values given for free surface charge $Q = 5 imes 10^{-6}\ \text{C}$ and bound induced charge $Q_{\text{ind}} = 4 imes 10^{-6}\ \text{C}$ into the equation: [cite: 183, 184, 814]
$$4 imes 10^{-6} = 5 imes 10^{-6} \left(1 - \frac{1}{K}\right)$$ [cite: 814]
$$\frac{4}{5} = 1 - \frac{1}{K} \implies \frac{1}{K} = 1 - \frac{4}{5} = \frac{1}{5}$$ [cite: 815]
$K = 5$ [cite: 815]
### Pattern Recognition
The fraction of charge induced on the dielectric face scales structurally as $\frac{K-1}{K}$[cite: 813, 815]. Observing a ratio of $4$ parts out of $5$ implies that the constant factor $K$ must equal $5$ directly[cite: 815].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q24 (2025)
The electric field in a region is given by $\vec{\mathrm{E}} = (2\hat{\mathrm{i}} + 4\hat{\mathrm{j}} + 6\hat{\mathrm{k}}) imes 10^{3}\mathrm{N} / mathrm{C}$ . The flux of the field through a rectangular surface parallel to x-z plane is $6.0\mathrm{Nm}^2\mathrm{C}^{-1}$ . The area of the surface is __________ $\mathrm{cm}^2$ . [cite: 195, 196]
### Related Formula
$$\phi = \vec{E} \cdot \vec{A}$$ [cite: 827]
### Core Logic
A surface aligned parallel to the $x\text{-}z$ plane possesses an area vector pointing completely orthogonal to it along the $y$-axis direction, meaning $\vec{A} = A\hat{j}$[cite: 196, 827]. Performing the dot product: [cite: 827]
$$\phi = \left[(2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3\right] \cdot (A\hat{j}) = 4 \times 10^3 A$$ [cite: 195, 827]
Given that the net flux magnitude is $6.0\ \text{Nm}^2\text{C}^{-1}$ [cite: 196]:
$$6 = 4 \times 10^3 A \implies A = \frac{6}{4 \times 10^3} = 1.5 \times 10^{-3}\ \text{m}^2$$ [cite: 828, 829]
Converting square meters to square centimeters ($1\ \text{m}^2 = 10^4\ \text{cm}^2$): [cite: 196, 830]
$$A = 1.5 \times 10^{-3} \times 10^4 = 15\ \text{cm}^2$$ [cite: 830]
### Pattern Recognition
Always focus exclusively on the specific field component matched to the surface orientation normal[cite: 827]. For an $x\text{-}z$ plane match, only the $\hat{j}$ coefficient creates flux[cite: 196, 827]. Do not miss the metric scale unit transition at the end ($m^2 \rightarrow cm^2$)[cite: 196, 830].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
