Electrostatics Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Electrostatics with step-by-step solutions covering Coulomb's law, capacitors, dipoles, Gauss's law and electric potential.
Solved JEE Main 2026 previous year questions on Electrostatics, worked out step by step.
Solutions are AI-generated and pending review.
Solution
The capacitor stays connected to the battery, so the voltage $V$ is constant.
Capacitance at separation $x$:
$$C = \frac{\epsilon_0 A}{x}$$Energy stored:
$$U = \frac{1}{2}CV^2 = \frac{\epsilon_0 A V^2}{2x}$$Differentiate with respect to time (with $\dfrac{dx}{dt} = v$):
$$\frac{dU}{dt} = \frac{\epsilon_0 A V^2}{2}\cdot\frac{d}{dt}\!\left(\frac{1}{x}\right) = -\frac{\epsilon_0 A V^2}{2}\cdot\frac{1}{x^2}\,v$$$$\frac{dU}{dt} \propto x^{-2}$$Hence $\alpha = -2$.
Answer: A ($-2$)
Solution
For a uniformly charged semicircular ring, the field at the centre is directed along the axis of symmetry. With linear charge density $\lambda = \dfrac{Q}{\pi R}$:
$$E = \frac{1}{4\pi\epsilon_0}\cdot\frac{2\lambda}{R} = \frac{2\lambda}{4\pi\epsilon_0 R}$$Substitute $\lambda = \dfrac{Q}{\pi R}$:
$$E = \frac{2Q}{4\pi\epsilon_0\,\pi R^2} = \frac{Q}{2\pi^2\epsilon_0 R^2}$$Solve for $Q$:
$$Q = 2\pi^2\epsilon_0 R^2 E = 2(3.14)^2(8.85\times10^{-12})(0.35)^2(100)$$$$Q \approx 2.14\times10^{-9}\ \text{C} = 2.14\ \text{nC}$$Answer: A (2.14)
Solution
Work done by the electric force:
$$W = q\int \vec{E}\cdot d\vec{l} = q\left[\int_{0}^{5} 2x\,dx + \int_{-2}^{1} 3y^2\,dy + \int_{-5}^{2} 4\,dz\right]$$Evaluate each integral:
$$\int_{0}^{5} 2x\,dx = \big[x^2\big]_0^5 = 25$$$$\int_{-2}^{1} 3y^2\,dy = \big[y^3\big]_{-2}^{1} = 1-(-8) = 9$$$$\int_{-5}^{2} 4\,dz = 4\big[z\big]_{-5}^{2} = 4(7) = 28$$Sum $= 25 + 9 + 28 = 62$. With $q = 3$ C:
$$W = 3\times 62 = 186\ \text{J}$$Answer: 186
Solution
When connected by a wire, both spheres reach the same potential:
$$V = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r_1} = \frac{1}{4\pi\epsilon_0}\frac{Q_2}{r_2}$$The surface field of a sphere is $E = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2} = \dfrac{V}{r}$.
Since $V$ is common:
$$\frac{E_{S_1}}{E_{S_2}} = \frac{V/r_1}{V/r_2} = \frac{r_2}{r_1} = \frac{18}{8} = \frac{9}{4}$$Answer: D ($\tfrac{9}{4}$)
Solution
A dipole of moment $p$ and moment of inertia $I$ oscillating in a field $E$ has angular frequency:
$$\omega = \sqrt{\frac{pE}{I}} \implies f \propto \sqrt{E}$$Initial field magnitude: $E_1 = E_0$.
Net field after adding $\vec{E}_2$:
$$\vec{E}_{net} = E_0\hat{x} + 2E_0\hat{y} + 2E_0\hat{z}$$$$|\vec{E}_{net}| = E_0\sqrt{1^2 + 2^2 + 2^2} = 3E_0$$Frequency ratio:
$$\frac{f_2}{f_1} = \sqrt{\frac{3E_0}{E_0}} = \sqrt{3} \approx 1.732$$Percentage change $= (\sqrt{3}-1)\times100 \approx 73\%$.
Answer: A (73%)
Solution
Assertion A is true: in electrostatic equilibrium the field inside a conductor is zero, so any net charge resides entirely on the surface — the interior holds no net charge.
Reason R is also true: a uniform electric field exists in the gap between capacitor plates, so a free charge placed there does experience a force and drifts.
However, R describes the behaviour of charges in the vacuum gap between plates, not the mechanism by which a conductor’s interior carries no net charge (which follows from $\vec{E}=0$ inside the conductor). So R does not correctly explain A.
Answer: B (Both A and R are true but R is NOT the correct explanation of A)
Solution
For the droplet to stay suspended, the electric force must balance gravity: $qE = mg$.
Mass of droplet ($\rho = 1500\ \text{kg/m}^3$, $r = 10^{-3}$ m):
$$m = \rho\cdot\frac{4}{3}\pi r^3 = 1500\cdot\frac{4}{3}\pi(10^{-3})^3 = 2\pi\times10^{-6}\ \text{kg}$$$$mg = 2\pi\times10^{-6}\times10 = 2\pi\times10^{-5}\ \text{N}$$Required field ($q = 5\times10^{-9}$ C):
$$E = \frac{mg}{q} = \frac{2\pi\times10^{-5}}{5\times10^{-9}} = 4\pi\times10^{3}\ \text{V/m}$$Potential difference ($d = \tfrac{12}{\pi}\times10^{-2}$ m):
$$\Delta V = E\,d = 4\pi\times10^{3}\times\frac{12}{\pi}\times10^{-2} = 480\ \text{V}$$The charge is positive, so the electric force must point up, meaning $\vec{E}$ points up (from bottom plate $B$ toward top plate $A$). Thus the bottom plate $B$ is at higher potential: $V_B - V_A = 480$ V. Only $V_A = 100$ V, $V_B = 580$ V satisfies this.
Answer: A ($100\,V$ and $580\,V$)
Solution
By Gauss’s law, flux depends only on the enclosed charge:
$$\Phi = \frac{Q_{enc}}{\epsilon_0}$$Sphere of radius 3 cm: encloses the charge at $x = 2$ cm (inside), but not the one at $x = 4$ cm (outside).
$$Q_{enc,1} = 8\,\mu C$$Sphere of radius 5 cm: encloses both charges ($x = 2$ cm and $x = 4$ cm).
$$Q_{enc,2} = 8 + (-2) = 6\,\mu C$$Ratio of fluxes:
$$\frac{\Phi_1}{\Phi_2} = \frac{8}{6} = \frac{4}{3}$$Answer: C ($4 : 3$)
Solution
The electric field is the negative gradient of potential:
$$\vec{E} = -\nabla V = -\frac{\partial V}{\partial x}\hat{i} - \frac{\partial V}{\partial y}\hat{j}$$With $V = 5(x^2 - y^2)$:
$$\frac{\partial V}{\partial x} = 10x, \qquad \frac{\partial V}{\partial y} = -10y$$So:
$$\vec{E} = -10x\,\hat{i} + 10y\,\hat{j}$$At $(2, 3)$:
$$\vec{E} = -10(2)\hat{i} + 10(3)\hat{j} = -20\hat{i} + 30\hat{j}\ \text{V/m}$$Answer: A ($-20\hat{i}+30\hat{j}$)
Solution
Initial energy ($C = 100$ pF, $V = 100$ V):
$$U_i = \frac{1}{2}CV^2 = \frac{1}{2}(100\times10^{-12})(100)^2 = 5\times10^{-7}\ \text{J}$$Charge is conserved: $Q = CV = 10^{-8}$ C. On contact the combined capacitance is $200$ pF, so the common potential becomes:
$$V' = \frac{Q}{C'} = \frac{10^{-8}}{200\times10^{-12}} = 50\ \text{V}$$Final energy:
$$U_f = \frac{1}{2}C'V'^2 = \frac{1}{2}(200\times10^{-12})(50)^2 = 2.5\times10^{-7}\ \text{J}$$Change in energy:
$$|\Delta U| = U_i - U_f = (5 - 2.5)\times10^{-7} = 2.5\times10^{-7}\ \text{J}$$So $\alpha = \dfrac{5}{2}$ (energy decreases by this amount).
Answer: B ($\tfrac{5}{2}$)
Solution
Displacement from $q_1$ to $q_2$:
$$\vec{r} = (1-2)\hat{i} + (1-3)\hat{j} + (1-3)\hat{k} = -\hat{i} - 2\hat{j} - 2\hat{k}$$$$|\vec{r}| = \sqrt{1 + 4 + 4} = 3$$Magnitude factor:
$$\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} = \frac{9\times10^9\times(3\times10^{-6})(-4\times10^{-6})}{3^2} = -1.2\times10^{-2}\ \text{N}$$Force on $q_2$ (directed along $\hat{r}$ from $q_1\to q_2$):
$$\vec{F} = \left(-1.2\times10^{-2}\right)\frac{\vec{r}}{|\vec{r}|} = \left(-1.2\times10^{-2}\right)\frac{(-\hat{i}-2\hat{j}-2\hat{k})}{3}$$$$\vec{F} = (4\times10^{-3})\hat{i} + (8\times10^{-3})\hat{j} + (8\times10^{-3})\hat{k}$$The charges are opposite, so $q_2$ is pulled toward $q_1$ along $(+\hat{i}+2\hat{j}+2\hat{k})$, as obtained.
$$\vec{F} = (4\hat{i} + 8\hat{j} + 8\hat{k})\times10^{-3}\ \text{N}$$Answer: B ($(4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}$)
Solution
Find the plate area from the resistance of the dielectric slab ($R = \rho\, d / A$):
$$A = \frac{\rho\, d}{R} = \frac{(10^{13})(0.885\times10^{-3})}{17.7\times10^{14}} = 5\times10^{-6}\ \text{m}^2$$Use the capacitance formula $C = \dfrac{\epsilon_0 \epsilon_r A}{d}$:
$$\epsilon_r = \frac{C\,d}{\epsilon_0 A} = \frac{(10^{-6})(0.885\times10^{-3})}{(8.85\times10^{-12})(5\times10^{-6})}$$$$\epsilon_r = \frac{8.85\times10^{-10}}{4.425\times10^{-17}} = 2\times10^{7}$$So $\alpha\times10^7 = 2\times10^7 \Rightarrow \alpha = 2$.
Answer: 2