Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 14.
Q40 (2026)
Two point charges 2q and q are placed at vertex A and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is :
{{IMG1}}
- $\frac{3q}{\varepsilon_0}$
- $\frac{q}{\varepsilon_0}$
- $\frac{3q}{2\varepsilon_0}$
- $\frac{3q}{4\varepsilon_0}$
### Related Formula
$$\phi = \frac{Q_{\text{enclosed}}}{\varepsilon_{0}}$$
### Core Logic
By Gauss's law, we consider the fraction of each charge that is geometrically "inside" the cube.
A charge at a vertex is shared equally among 8 adjoining cubes. So its contribution to this specific cube is $1/8$.
A charge at the center of a face is shared equally between 2 adjoining cubes. So its contribution to this specific cube is $1/2$.
### Step 1: Calculate Enclosed Charge
Charge at vertex A = $2q$. Inside portion = $\frac{1}{8}(2q) = \frac{q}{4}$.
Charge at face center = $q$. Inside portion = $\frac{1}{2}(q) = \frac{q}{2}$.
$$Q_{\text{in}} = \frac{q}{4} + \frac{q}{2} = \frac{3q}{4}$$
### Step 2: Final Flux Calculation
$$\phi = \frac{Q_{\text{in}}}{\varepsilon_{0}} = \frac{\frac{3q}{4}}{\varepsilon_{0}} = \frac{3q}{4\varepsilon_{0}}$$
### Pattern Recognition
Sees: "flux through cube" + "charges on boundaries" → strictly evaluate solid angle fractions. Vertex = 1/8, Edge = 1/4, Face = 1/2.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q50 (2026)
The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_{1}=2$, $K_{2}=3$, and $K_{3}=5$ (as shown in the figure). If new capacitance is $\frac{n}{3}C$ then the value of n is ____.
{{IMG1}}
### Related Formula
$$C = \frac{\varepsilon_{0}A}{d}$$
$$C_{\text{dielectric}} = K \frac{\varepsilon_{0}A'}{d'}$$
Series: $C_{\text{eq}} = \left(\frac{1}{C_{1}} + \frac{1}{C_{2}}\right)^{-1}$
Parallel: $C_{\text{eq}} = C_{1} + C_{2}$
### Core Logic
The original capacitance is $C = \frac{\varepsilon_{0}A}{d}$. We mentally split the system into 4 distinct block capacitors based on area and thickness splits.
Upper half contains $K_{1}$, acting on full Area but half depth. It can be split into two parallel halves to match the lower blocks.
Left branch: Top block $C_{1}$ (area $A/2$, depth $d/2$, $K=2$) in series with Bottom block $C_{2}$ (area $A/2$, depth $d/2$, $K=3$).
Right branch: Top block $C_{4}$ (area $A/2$, depth $d/2$, $K=2$) in series with Bottom block $C_{3}$ (area $A/2$, depth $d/2$, $K=5$).
### Step 1: Calculate Block Capacitances
$$C_{1} = \frac{2 \varepsilon_{0} (A/2)}{d/2} = \frac{2 \varepsilon_{0} A}{d} = 2C$$
$$C_{2} = \frac{3 \varepsilon_{0} (A/2)}{d/2} = \frac{3 \varepsilon_{0} A}{d} = 3C$$
$$C_{4} = \frac{2 \varepsilon_{0} (A/2)}{d/2} = \frac{2 \varepsilon_{0} A}{d} = 2C$$
$$C_{3} = \frac{5 \varepsilon_{0} (A/2)}{d/2} = \frac{5 \varepsilon_{0} A}{d} = 5C$$
### Step 2: Resolve Left and Right Series Combinations
Wait, the provided solution assigns block $C_1 = 5C$ and $C_2=2C$. It splits it differently: upper block $C_{upper} = \frac{2 \varepsilon_{0} A}{d/2} = 4C$. Lower left $C_{LL} = \frac{3 \varepsilon_{0} A/2}{d/2} = 3C$. Lower right $C_{LR} = \frac{5 \varepsilon_{0} A/2}{d/2} = 5C$.
Then Lower parallel $C_{L} = 3C + 5C = 8C$.
Then $C_{eq} = \frac{4C \times 8C}{4C + 8C} = \frac{32}{12}C = \frac{8}{3}C$.
Let's check the PDF solution logic: the solution calculates $C_1 = 5C, C_2 = 2C$ in series. It seems it mismatched K values to the variable names but the math output is $C_{eq} = \frac{92}{35}C \approx 2.62C = \frac{7.9}{3}C \implies n \approx 8$.
Wait, the solution logic says: $C_{eq} = \frac{92}{35}C \approx 2.62 C$.
The correct pure method:
Upper block $K_1=2$. $C_1 = \frac{2 \varepsilon_{0} A}{d/2} = 4C$.
Bottom blocks are parallel: $C_2 = \frac{3 \varepsilon_{0} A/2}{d/2} = 3C$, $C_3 = \frac{5 \varepsilon_{0} A/2}{d/2} = 5C$.
Equivalent bottom $= 3C + 5C = 8C$.
Total $C_{eq} = \frac{4C \cdot 8C}{4C + 8C} = \frac{32}{12}C = \frac{8}{3}C$.
So exact $n = 8$.
The PDF solution makes an arithmetic convolution splitting it vertically fully into two branches but correctly arrives at approximately 8. We will use the direct correct layout that yields exactly 8.
### Step 3: Final Capacitance
$$C_{\text{eq}} = \frac{8}{3} C$$
We are given $C_{\text{eq}} = \frac{n}{3} C$.
Comparing both, we get $n = 8$.
### Pattern Recognition
Sees: "composite dielectrics with splits" → Treat series divisions (depth cuts) as series capacitors and parallel divisions (area cuts) as parallel capacitors. Horizontal boundary = Series. Vertical boundary = Parallel.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q31 (2026)
Two shorts dipoles $(A, B)$ , A having charges $\pm 2\mu C$ and length 1 cm and B having charges $\pm 4\mu C$ and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point P, equi-distant from the centres of both dipoles is ____ N/C.
{{IMG1}}
- $\frac{9}{16}\sqrt{2} \times 10^{5}$
- $4.5 \sqrt{2} \times 10^{4}$
- $9 \sqrt{2} \times 10^{4}$
- $\frac{9}{16}\sqrt{2} \times 10^{4}$
### Related Formula
Electric field due to a short dipole on its axial line:
$$E_{\text{axial}} = \frac{2Kp}{r^3}$$
Electric field due to a short dipole on its equatorial line:
$$E_{\text{equatorial}} = -\frac{Kp}{r^3}$$
### Core Logic
{{SOL_IMG1}}
Point P is at a distance of $r = 40 \, \mathrm{cm} = 0.4 \, \mathrm{m}$ from the centre of both dipoles.
For dipole A, point P lies on the axial line.
For dipole B, point P lies on the equatorial line.
Dipole moments:
$$P_1 = q \times d = 2 \times 10^{-6} \times 10^{-2} = 2 \times 10^{-8} \, \mathrm{C\cdot m}$$
$$P_2 = q \times d = 4 \times 10^{-6} \times 10^{-2} = 4 \times 10^{-8} \, \mathrm{C\cdot m}$$
### Step 1: Calculate Field Components
$$\vec{E}_1 = -\frac{2KP_1}{r^3} \hat{i}$$
Wait, the orientation in the figure sets the directions. From the diagram, A is oriented left-to-right (minus to plus) so $\vec{E}_1$ on its axis is along $+\hat{i}$ or $-\hat{i}$ depending on precise placement. Based on standard convention:
$$\vec{E}_1 = \frac{2KP_1}{r^3} \hat{i}$$
$$\vec{E}_2 = -\frac{KP_2}{r^3} \hat{j}$$ (Equatorial field points opposite to dipole vector)
Net Field vector based on solution structure:
$$\vec{E}_{\text{net}} = \frac{2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{(0.4)^3} \hat{i} - \frac{9 \times 10^9 \times 4 \times 10^{-8}}{(0.4)^3} \hat{j}$$
### Step 2: Find Magnitude of Net Field
$$\vec{E}_{\text{net}} = \frac{9 \times 10^9 \times 10^{-8}}{(0.4)^3} \left[ (2 \times 2)\hat{i} - 4\hat{j} \right]$$
Wait, the solution simplifies it as:
$$\vec{E}_{\text{net}} = \frac{9 \times 10^9 \times 4 \times 10^{-8}}{(0.4)^3} [ \hat{i} - \hat{j} ]$$
Magnitude:
$$|\vec{E}_{\text{net}}| = \frac{360}{(0.064)} \sqrt{1^2 + (-1)^2} = \frac{360 \sqrt{2}}{0.064}$$
$$|\vec{E}_{\text{net}}| = \frac{360 \times 10^3}{64} \sqrt{2} = \frac{90}{16} \times 10^3 \sqrt{2} = \frac{9}{16} \sqrt{2} \times 10^4 \, \mathrm{N/C}$$
### Pattern Recognition
Identify the relationship of the observation point to each dipole (axial vs equatorial). The axial field has a factor of 2, while the equatorial does not. Vector addition is mandatory since the fields are perpendicular.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q32 (2026)
Two charges $7\mu C$ and $-2\mu C$ are placed at $(-9,0,0)$ cm and $(9,0,0)$ cm respectively in an external field $E=\frac{A}{r^{2}}\hat{r}$ , where $A=9\times10^{5}\mathrm{N/C.m^{2}}$ . Considering the potential at infinity is 0, the electrostatic energy of the configuration is ____ J.
- $1.4$
- $-90.7$
- $49.3$
- $24.3$
### Related Formula
$$dV = -\vec{E} \cdot d\vec{r}$$
$$U_{\text{total}} = U_{\text{self}} + U_{\text{interaction}} = q_1 V_1 + q_2 V_2 + \frac{kq_1 q_2}{r_{12}}$$
### Core Logic
First, derive the potential function $V(r)$ from the given external electric field.
$$V(r) = - \int_{\infty}^r \frac{A}{r^2} \, dr = - \left[ -\frac{A}{r} \right]_{\infty}^r = \frac{A}{r}$$
Both charges are at distance $r = 9 \, \mathrm{cm} = 0.09 \, \mathrm{m}$ from the origin (where the field points radiate from).
### Step 1: Calculate Energy of Individual Charges in External Field
Energy of $q_1$ in the external field:
$$U_1 = q_1 V_1 = (7 \times 10^{-6}) \left( \frac{A}{9 \times 10^{-2}} \right)$$
Energy of $q_2$ in the external field:
$$U_2 = q_2 V_2 = (-2 \times 10^{-6}) \left( \frac{A}{9 \times 10^{-2}} \right)$$
Sum of self energies:
$$U_{\text{self}} = \frac{5 \times 10^{-6} \times 9 \times 10^5}{9 \times 10^{-2}} = \frac{4.5}{0.09} = 50 \, \mathrm{J}$$
### Step 2: Calculate Interaction Energy Between Charges
Distance between charges $r_{12} = 9 - (-9) = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m}$.
$$U_{\text{interaction}} = \frac{k q_1 q_2}{r_{12}}$$
$$U_{\text{interaction}} = \frac{9 \times 10^9 \times (7 \times 10^{-6}) \times (-2 \times 10^{-6})}{18 \times 10^{-2}}$$
$$U_{\text{interaction}} = - \frac{9 \times 14 \times 10^{-3}}{0.18} = -0.7 \, \mathrm{J}$$
### Step 3: Total Energy
$$U_{\text{total}} = U_{\text{self}} + U_{\text{interaction}}$$
$$U_{\text{total}} = 50 - 0.7 = 49.3 \, \mathrm{J}$$
### Pattern Recognition
When an external field is present, total energy is the sum of the potential energies of each charge independently in that external field, plus the mutual interaction energy between the charges themselves.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
Q34 (2026)
A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws 25% more charge from the battery. The dielectric constant of mica is ____.
- $2.5$
- $2.0$
- $1.5$
- $1.0$
### Related Formula
$$C = \frac{\epsilon_0 A}{d}$$
$$C_{eq} = \frac{\epsilon_0 A}{(d - t) + \frac{t}{K}}$$
$Q = CV$
### Core Logic
{{SOL_IMG1}}
{{SOL_IMG2}}
{{SOL_IMG3}}
Initially, capacitance is $C = \frac{\epsilon_0 A}{5}$. The battery maintains constant voltage $V$.
Initial charge $Q_1 = CV$.
When a $2 \, \mathrm{mm}$ mica sheet is introduced, it acts as two capacitors in series: air gap ($3 \, \mathrm{mm}$) and mica gap ($2 \, \mathrm{mm}$).
New charge $Q_2 = 1.25 Q_1$ (since it draws 25% more).
Thus, $C_{eq} = 1.25 C$.
### Step 1: Calculate Equivalent Capacitance
$$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{\frac{\epsilon_0 A}{3} \times \frac{K \epsilon_0 A}{2}}{\frac{\epsilon_0 A}{3} + \frac{K \epsilon_0 A}{2}}$$
$$C_{eq} = \frac{\epsilon_0 A}{\frac{3}{1} + \frac{2}{K}} = \frac{K \epsilon_0 A}{3K + 2}$$
### Step 2: Equate to New Charge Requirement
$$1.25 \times \frac{\epsilon_0 A}{5} = \frac{K \epsilon_0 A}{3K + 2}$$
$$0.25 = \frac{K}{3K + 2}$$
$$0.25 (3K + 2) = K$$
$0.75K + 0.5 = K$
$0.25K = 0.5$
$K = 2$
### Pattern Recognition
When a dielectric slab of thickness $t$ is inserted, equivalent capacitance is directly modeled as $C' = \frac{\epsilon_0 A}{(d - t) + t/K}$. Constant voltage implies $Q \propto C$, directly yielding the ratio equation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatics
