Electrostatics Previous Year Questions — JEE Main Physics
79 past-year Electrostatics questions from JEE Main (Physics) — page 3.
Q7 (2025)
An infinitely long wire has uniform linear charge density $\lambda = 2\mathrm{~nC/m}$. The net flux through a Gaussian cube of side length $\sqrt{3}\mathrm{~cm}$, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be $x\mathrm{~N}\cdot\mathrm{m}^{2}\cdot\mathrm{C}^{-1}$, where $x$ is:
[Neglect any edge effects and use $\frac{1}{4\pi\varepsilon_0} = 9\times 10^9$ SI units]
- $0.72\pi$
- $1.44\pi$
- $6.48\pi$
- $2.16\pi$
### Related Formula
$$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}$$
$$q_{\text{enc}} = \lambda \cdot L_{\text{enclosed}}$$
$$\frac{1}{\varepsilon_0} = 4\pi \left(9 \times 10^9\right) = 36\pi \times 10^9\mathrm{~N\cdot m^2\cdot C^{-2}}$$
### Core Logic
The two corners of the cube that are maximally displaced from each other represent the body diagonal of the cube.
- Side length of the cube, $a = sqrt{3}\mathrm{~cm} = \sqrt{3} \times 10^{-2}\mathrm{~m}$
- Length of the body diagonal (length of the wire enclosed inside the cube):
$$L_{\text{enclosed}} = \sqrt{3} a = \sqrt{3} \left(\sqrt{3} \times 10^{-2}\mathrm{~m}\right) = 3 \times 10^{-2}\mathrm{~m} = 3\mathrm{~cm}$$
Now, find the enclosed charge $q_{\text{enc}}$:
$$q_{\text{enc}} = \lambda \cdot L_{\text{enclosed}} = \left(2 \times 10^{-9}\mathrm{~C/m}\right) \times \left(3 \times 10^{-2}\mathrm{~m}\right) = 6 \times 10^{-11}\mathrm{~C}$$
### Step 1: Net Flux Computation
Using Gauss's Law:
$$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0} = 6 \times 10^{-11} \times \left(36\pi \times 10^9\right)$$
$$\Phi = 216\pi \times 10^{-2} = 2.16\pi\mathrm{~N\cdot m^2\cdot C^{-1}}$$
Thus, comparing with $x\mathrm{~N\cdot m^2\cdot C^{-1}}$ yields:
$x = 2.16\pi$
### Pattern Recognition
Sees: "Wire passing through maximally displaced corners of a cube" → The length inside is the body diagonal $= \sqrt{3} a$.
Shortcut: Convert units carefully. Since $a = \sqrt{3}\mathrm{~cm}$, the body diagonal becomes exactly $3\mathrm{~cm}$. Multiplying linear charge density directly gives the charge inside. Using $\frac{1}{\varepsilon_0} = 36\pi \times 10^9$ ensures $\pi$ is easily kept in the final answer. ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatics
Q9 (2025)
Two metal spheres of radius $R$ and $3R$ have same surface charge density $\sigma$. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes $\sigma_{1}$ and $\sigma_{2}$, respectively. The ratio $\frac{\sigma_{1}}{\sigma_{2}}$ is:
- $\frac{1}{9}$
- $9$
- $\frac{1}{3}$
- $3$
### Related Formula
$$V = \frac{\sigma r}{\varepsilon_0}$$
where,
$V$ = electrostatic potential of a conducting sphere
$\sigma$ = surface charge density
$r$ = radius of the sphere
### Core Logic
For any conducting sphere, the potential on its surface is related to its surface charge density by:
$$V = \frac{k Q}{r} = \frac{1}{4\pi\varepsilon_0} \frac{\sigma \left(4\pi r^2\right)}{r} = \frac{\sigma r}{\varepsilon_0}$$
When the two spheres of radii $r_1 = R$ and $r_2 = 3R$ are brought into contact, charge flows between them until they reach an identical electric potential:
$V_1 = V_2$
### Step 1: Ratio Calculation
Equate the potentials of the two spheres after separation:
$$\frac{\sigma_1 r_1}{\varepsilon_0} = \frac{\sigma_2 r_2}{\varepsilon_0}$$
$$\sigma_1 R = \sigma_2 (3R) \implies \frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3$$
### Pattern Recognition
Sees: "Conducting spheres brought in contact" → Electric potentials become equal: $V_1 = V_2$.
Shortcut: Since $V \propto \sigma r$, equal potential directly implies $\sigma_1 r_1 = \sigma_2 r_2$. Thus, the ratio of final densities is simply the inverse ratio of their radii: $\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1} = \frac{3}{1} = 3$. This bypasses computing the individual final charges entirely! ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatics
Q23 (2025)
Space between the plates of a parallel plate capacitor of plate area $4\mathrm{~cm}^{2}$ and separation of (d) $1.77\mathrm{~mm}$, is filled with uniform dielectric materials with dielectric constants ($3$ and $5$) as shown in figure. Another capacitor of capacitance $7.5\mathrm{~pF}$ is connected in parallel with it. The effective capacitance of this combination is ________ $\mathrm{~pF}$.
{{IMG1}}
(Given $\varepsilon_0 = 8.85\times 10^{-12}\mathrm{~F/m}$)
### Related Formula
$$C = k \frac{\varepsilon_0 A}{t}$$
$$\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}$$
$$C_{\text{parallel}} = C_{\text{eq}} + C_p$$
### Core Logic
The dielectrics $k_1 = 5$ and $k_2 = 3$ split the capacitor separation horizontally into two layers, each of thickness $\frac{d}{2}$. Thus, they act as two capacitors connected in **series**:
- Capacitor 1: $C_1 = k_1 \frac{\varepsilon_0 A}{d/2} = 5 \times \frac{2 \varepsilon_0 A}{d} = 10 \frac{\varepsilon_0 A}{d}$
- Capacitor 2: $C_2 = k_2 \frac{\varepsilon_0 A}{d/2} = 3 \times \frac{2 \varepsilon_0 A}{d} = 6 \frac{\varepsilon_0 A}{d}$
### Step 1: Compute Base Capacitance Factor
First, find the term $\frac{\varepsilon_0 A}{d}$ in SI units:
- $A = 4\mathrm{~cm}^2 = 4 \times 10^{-4}\mathrm{~m}^2$
- $d = 1.77\mathrm{~mm} = 1.77 \times 10^{-3}\mathrm{~m}$
- $\varepsilon_0 = 8.85 \times 10^{-12}\mathrm{~F/m}$
$$\frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}} = \frac{35.4 \times 10^{-16}}{1.77 \times 10^{-3}} = 20 \times 10^{-13}\mathrm{~F} = 2\mathrm{~pF}$$
Now find $C_1$ and $C_2$:
- $C_1 = 10 \times 2\mathrm{~pF} = 20\mathrm{~pF}$
- $C_2 = 6 \times 2\mathrm{~pF} = 12\mathrm{~pF}$
Calculate their series equivalent $C_{\text{eq}}$:
$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{20 \times 12}{20 + 12} = \frac{240}{32} = 7.5\mathrm{~pF}$$
### Step 2: Add Parallel Capacitor
Another capacitor of $C_p = 7.5\mathrm{~pF}$ is connected in parallel with the combination:
$$C_{\text{final}} = C_{\text{eq}} + C_p = 7.5\mathrm{~pF} + 7.5\mathrm{~pF} = 15\mathrm{~pF}$$
### Pattern Recognition
Sees: Dielectric boundary parallel to the plates → Series capacitors.
Trap: Don't treat horizontally split layers as parallel; split in distance $d$ means series, while split in area $A$ means parallel.
Shortcut: Notice $\frac{35.4}{1.77}$ is exactly $20$. This makes the numerical calculations incredibly clean! ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatics
Q3 (2025)
An electric dipole is placed at a distance of $2\mathrm{~cm}$ from an infinite plane sheet having positive charge density $\sigma_0$. Choose the correct option from the following.
{{IMG1}}
- $\text{Torque on dipole is zero and net force is directed away from the sheet.}$
- $\text{Torque on dipole is zero and net force acts towards the sheet.}$
- $\text{Potential energy of dipole is minimum and torque is zero.}$
- $\text{Potential energy and torque both are maximum}$
### Related Formula
$$E = \frac{\sigma_0}{2\epsilon_0}$$
$$\vec{\tau} = \vec{p} \times \vec{E}$$
$$U = -\vec{p} \cdot \vec{E}$$
### Core Logic
An infinite plane sheet produces a uniform electric field $\vec{E}$ directed normally away from the sheet.
As shown in the image layout, the dipole moment vector $\vec{p}$ (pointing from $-q$ to $+q$) is oriented parallel to the electric field vectors $\vec{E}$:
$$\theta = 0^{\circ}$$
1. **Torque evaluation**:
$$\tau = pE \sin(0^{\circ}) = 0$$
2. **Potential Energy evaluation**:
$$U = -pE \cos(0^{\circ}) = -pE \quad (\text{Minimum})$$
3. **Net Force evaluation**:
Since the electric field is uniform, the force on $+q$ balances the force on $-q$, meaning $F_{\text{net}} = 0$.
### Pattern Recognition
When a dipole aligns perfectly with a uniform electric field ($\vec{p} \parallel \vec{E}$), it reaches stable equilibrium. Stable equilibrium fundamentally means minimum potential energy ($U = -pE$) and zero torque.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatics
Q5 (2025)
A point charge causes an electric flux of $-2 \times 10^{4} \mathrm{~Nm}^{2}\mathrm{C}^{-1}$ to pass through a spherical Gaussian surface of $8.0\mathrm{~cm}$ radius, centred on the charge. The value of the point charge is :
$(\mathrm{Given } \epsilon_{0} = 8.85 \times 10^{-12} \mathrm{~C}^{2}\mathrm{N}^{-1}\mathrm{m}^{-2})$
- $-17.7 \times 10^{-8} \mathrm{~C}$
- $-15.7 \times 10^{-8} \mathrm{~C}$
- $17.7 \times 10^{-8} \mathrm{~C}$
- $15.7 \times 10^{-8} \mathrm{~C}$
### Related Formula
$$\phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$$
where,
$\phi$ = net electric flux through the closed surface
$q_{\text{enclosed}}$ = net charge enclosed by the surface
$\epsilon_0$ = permittivity of free space
### Core Logic
According to Gauss's Law, the total electric flux through a closed surface depends only on the charge enclosed inside it, completely independent of the radius of the surface.
Rearranging the formula to solve for $q$:
$$q = \phi \cdot \epsilon_0$$
Substitute the given values:
$$q = (-2 \times 10^{4} \mathrm{~Nm}^2\mathrm{C}^{-1}) \times (8.85 \times 10^{-12} \mathrm{~C}^2\mathrm{N}^{-1}\mathrm{m}^{-2})$$
$$q = -17.7 \times 10^{-8} \mathrm{~C}$$
### Pattern Recognition
Distractor alert: The radius ($8.0\mathrm{~cm}$) is extra data meant to mislead. Gauss's flux depends entirely on the magnitude of the enclosed charge, not the physical surface area configuration.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electrostatics
