Electrostatics JEE Main previous year questions with solutions

5 solved JEE Main questions on Electrostatics, free to read — no sign-in needed. The full chapter has 299 questions; sign in to attempt the remaining 294 in the exam simulator.

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1. Q1JEE Main 2026 (08 Apr, Shift 2)Electric Charge and Coulomb's law
   Two point charges $q_1=3\,\mu C$ and $q_2=-4\,\mu C$ are placed at points $(2\hat{i}+3\hat{j}+3\hat{k})$ and $(\hat{i}+\hat{j}+\hat{k})$ respectively. Force on charge $q_2$ is ________ N. $\left(\text{Take } \dfrac{1}{4\pi\epsilon_0} = 9\times 10^9 \text{ SI Units}\right)$
   1. A.$(12\hat{i}+24\hat{j}+24\hat{k})\times 10^{-3}$
   2. B.$(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}$
   3. C.$(3\hat{i}+6\hat{j}+6\hat{k})\times 10^{-3}$
   4. D.$(-4\hat{i}-8\hat{j}-8\hat{k})\times 10^{-3}$
   Show answer & solution

   Answer: (B)

   Position vector of $q_1$, $\vec{r}_1 = 2\hat{i} + 3\hat{j} + 3\hat{k}$ Position vector of $q_2$, $\vec{r}_2 = \hat{i} + \hat{j} + \hat{k}$ Vector from $q_1$ to $q_2$ is $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1 = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} + 3\hat{k}) = -\hat{i} - 2\hat{j} - 2\hat{k}$ Magnitude $|\vec{r}_{21}| = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{9} = 3$ Force on $q_2$ due to $q_1$ is given by Coulomb's law in vector form: $\vec{F}_{21} = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{|\vec{r}_{21}|^3} \vec{r}_{21}$ Substituting the given values: $\vec{F}_{21} = \dfrac{9 \times 10^9 \times (3 \times 10^{-6}) \times (-4 \times 10^{-6})}{3^3} (-\hat{i} - 2\hat{j} - 2\hat{k})$ $\vec{F}_{21} = \dfrac{-108 \times 10^{-3}}{27} (-\hat{i} - 2\hat{j} - 2\hat{k})$ $\vec{F}_{21} = -4 \times 10^{-3} (-\hat{i} - 2\hat{j} - 2\hat{k})$ $\vec{F}_{21} = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3} \text{ N}$ Answer: $(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}$
2. Q2JEE Main 2025 (07 Apr, Shift 2)Electric Dipole
   <p>A dipole with two electric charges of $2 \mu \mathrm{C}$ magnitude each, with separation distance $0.5 \mu \mathrm{~m},$ 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 :</p>
   1. A.$5 \times 10^{-9} \mathrm{Nm}$
   2. B.$5 \times 10^{-3} \mathrm{Nm}$
   3. C.$2.5 \times 10^{-12} \mathrm{Nm}$
   4. D.<p>$2.5 \times 10^{-9} \mathrm{Nm}$</p>
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   Answer: (A)

   <p>$\begin{aligned} & \mathrm{E}=\frac{\mathrm{v}}{\mathrm{~d}}=\frac{5}{5 \times 10^{-4}}=10^4 \mathrm{~V} / \mathrm{m} \\ & \tau=\mathrm{PE} \sin \theta \end{aligned}$ Where $\mathrm{P}=\mathrm{qa}=2 \times 10^{-6} \times 5 \times 10^{-7}$ $=1 \times 10^{-12} \mathrm{C}-\mathrm{m}$ $\tau=1 \times 10^{-12} \times 10^4 \times \frac{1}{2}=5 \times 10^{-9} \mathrm{~N}-\mathrm{m}$</p>
3. Q3JEE Main 2024 (29 Jan, Shift 1)Electric Field and Electric Field Lines
   An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet $S$ having surface charge density $+\sigma$. The electron at $t=0$ is at a distance of $1m$ from $S$ and has a speed of $1m{s}^{-1}$. The maximum value of $\sigma$, if the electron strikes $S$ at $t=1s$ is $\alpha \left[\dfrac{m{ϵ}_{0}}{e}\right]\dfrac{C}{{m}^{2}}$. The value of $\alpha$ is _____.
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   Answer: 8

   The value of the electric field due to the uniformly charged plane sheet is given by $E=\dfrac{\sigma }{2{\epsilon }_{0}}...\left(1\right)$ Thus, the force on the electron is given by $\begin{matrix}F & = & -eE \\ & = & -\dfrac{\sigma e}{2{\epsilon }_{0}}...\left(2\right)\end{matrix}$ So, the acceleration of the electron is given by $a=-\dfrac{\sigma e}{2{\epsilon }_{0}m}$ Given, $u=1m{s}^{-1}$, $t=1s$ and $S=-1m$ Using the equation of motion, it can be written that $-1=1\times 1-\dfrac{1}{2}\times \dfrac{\sigma e}{2{\epsilon }_{0}m}\times (1{)}^{2} \Rightarrow \sigma =8\dfrac{{\epsilon }_{0}m}{e}C{m}^{-2}$ $∴\alpha =8$.
4. Q4JEE Main 2023 (08 Apr, Shift 2)Electric Potential and Potential Energy
   Electric potential at a point $P$ due to a point charge of $5\times {10}^{-9}C$ is $50V$. The distance of $P$ from the point charge is: (Assume, $\dfrac{1}{4\pi {\epsilon }_{0}}=9\times {10}^{9}N{m}^{2}{C}^{-2}$)
   1. A.$9cm$
   2. B.$3cm$
   3. C.$0.9cm$
   4. D.$90cm$
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   Answer: (D)

   The electric potential for a point charge is given by $V=\dfrac{Q}{4\pi {\epsilon }_{0}r}$ The data given is $V=50V Q=5\times {10}^{-9}C$ So, the distance of the point is $r=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{Q}{V}$ $\Rightarrow r=9\times {10}^{9}\times \dfrac{5\times {10}^{-9}}{50}$ $\Rightarrow r=0.9m=90cm$
5. Q5JEE Main 2022 (25 Jul, Shift 1)Electric Flux and Gauss Law
   The volume charge density of a sphere of radius $6m$ is $2\mu C{cm}^{-3}$. The number of lines of force per unit surface area coming out from the surface of the sphere is _____ $\times {10}^{10}N{C}^{-1}.$ [Given : Permittivity of vacuum ${ϵ}_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}-{m}^{-2}$]
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   Answer: 45

   Given here, $\rho =2\mu C{cm}^{-3}$ and $R=6m$. Number of lines of force per unit area $=$ electric field at surface$=\dfrac{KQ}{{R}^{2}}$ $E=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\rho \dfrac{4}{3}\pi {R}^{3}}{{R}^{2}}$ $E=\dfrac{\rho R}{3{ϵ}_{0}}=\dfrac{2\times 6}{3\times 8.85\times {10}^{-12}}=0.45\times {10}^{12}N{C}^{-1}$ $=45\times {10}^{10}N{C}^{-1}$