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 306 questions; sign in to attempt the remaining 301 in the exam simulator.

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.$(-4\hat{i}-8\hat{j}-8\hat{k})\times 10^{-3}$
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   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 (30 Jan, Shift 2)Electric Field and Electric Field Lines
   A particle of charge $-q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+\lambda$. Then time period will be given as: (Consider $k$ as Coulomb's constant)
   1. A.${T}^{2}=\dfrac{4{\pi }^{2}m}{2k\lambda q}{r}^{3}$
   2. B.$T=2\pi r\sqrt{\dfrac{m}{2k\lambda q}}$
   3. C.$T=\dfrac{1}{2\pi r}\sqrt{\dfrac{m}{2k\lambda q}}$
   4. D.$T=\dfrac{1}{2\pi }\sqrt{\dfrac{2k\lambda q}{m}}$
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   Answer: (B)

   Attractive electrostatic force due to line charge on the charged particle will provide the required centripetal force. Therefore, $q\left(\dfrac{2k\lambda }{r}\right)=m{\omega }^{2}r$ $\Rightarrow {\omega }^{2}=\dfrac{2k\lambda q}{m{r}^{2}}$ $\Rightarrow {\left(\dfrac{2\pi }{T}\right)}^{2}=\dfrac{2k\lambda q}{m{r}^{2}}$ $\Rightarrow T=2\pi r\sqrt{\dfrac{m}{2k\lambda q}}$
4. Q4JEE Main 2023 (11 Apr, Shift 2)Electric Potential and Potential Energy
   If $V$ is the gravitational potential due to sphere of uniform density on its surface, then its value at the centre of sphere will be:
   1. A.$\dfrac{4}{3}V$
   2. B.$\dfrac{V}{2}$
   3. C.$V$
   4. D.$\dfrac{3V}{2}$
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   Answer: (D)

   The magnitude of gravitational potential of a solid sphere at a distance $r$ from the centre is $V=\dfrac{-GM}{2{R}^{3}}(3{R}^{2}-{r}^{2})$ At $r=R$, $V=\dfrac{-GM}{R}$. At $r=0$, $V'=\dfrac{-GM}{2{R}^{3}}(3{R}^{2})=\dfrac{-3GM}{2R}$ Clearly, $V'=\dfrac{3V}{2}$.
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 ${ϵ}_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}-{m}^{-2}$]
   Show answer & solution

   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$=\dfrac{KQ}{{R}^{2}}$ $E=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\rho \dfrac{4}{3}\pi {R}^{3}}{{R}^{2}}$ $E=\dfrac{\rho R}{3{ϵ}_{0}}=\dfrac{2\times 6}{3\times 8.85\times {10}^{-12}}=0.45\times {10}^{12}N{C}^{-1}$ $=45\times {10}^{10}N{C}^{-1}$