Electrostatics JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Electrostatics, free to read — no sign-in needed. The full chapter has 62 questions; sign in to attempt the remaining 57 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Consider an electric dipole comprising two charges $+q$ and $-q$ each with mass $m$, separated by a fixed distance $d$ and initially at rest with its dipole moment pointing along $\hat{i}$. A uniform electric field $E\hat{j}$ is turned on at time $t = 0$ and it is turned off at $t = t_f$, when the dipole moment makes an angle $\theta_f$ with $\hat{i}$. Neglecting any sources of energy loss, correct option(s) is/are:
   1. A.The center of mass of the dipole is deflected towards $\hat{j}$ in the presence of the field.
   2. B.If the magnitude of the final angular velocity $\omega_f = \sqrt{\dfrac{2qE}{md}}$, then $\theta_f = \dfrac{\pi}{6}$.
   3. C.If $\theta_f = \pi/3$, then the change in kinetic energy of the dipole is given by $2\sqrt{3}\, qEd$.
   4. D.For $\theta_f = \pi/4$, the dipole rotates around its center of mass with a constant angular velocity after $t \gt t_f$.
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   Answer: B,D

   The net force on the dipole in a uniform electric field is zero, as the forces on the two charges are equal and opposite ($qE\hat{j}$ and $-qE\hat{j}$). Therefore, the center of mass remains at rest and is not deflected. Option (A) is incorrect. The moment of inertia of the dipole about its center of mass is $I = m\left(\dfrac{d}{2}\right)^2 + m\left(\dfrac{d}{2}\right)^2 = \dfrac{md^2}{2}$. The torque on the dipole is $\vec{\tau} = \vec{p} \times \vec{E}$. The angle between the dipole moment $\vec{p}$ and the electric field $\vec{E}$ is $90^{\circ} - \theta$. The magnitude of the torque is $\tau = pE \sin(90^{\circ} - \theta) = qdE \cos\theta$. The work done by the electric field as the dipole rotates from $\theta = 0$ to $\theta = \theta_f$ is equal to the change in kinetic energy: $\Delta K = \int_{0}^{\theta_f} \tau d\theta = \int_{0}^{\theta_f} qdE \cos\theta d\theta = qdE \sin\theta_f$ Equating this to the final kinetic energy $\dfrac{1}{2}I\omega_f^2$: $\dfrac{1}{2} \left(\dfrac{md^2}{2}\right) \omega_f^2 = qdE \sin\theta_f$ $\dfrac{md^2}{4} \omega_f^2 = qdE \sin\theta_f$ For option (B), substituting $\omega_f = \sqrt{\dfrac{2qE}{md}}$: $\dfrac{md^2}{4} \left(\dfrac{2qE}{md}\right) = qdE \sin\theta_f$ $\dfrac{qdE}{2} = qdE \sin\theta_f \Rightarrow \sin\theta_f = \dfrac{1}{2} \Rightarrow \theta_f = \dfrac{\pi}{6}$ Thus, option (B) is correct. For option (C), if $\theta_f = \dfrac{\pi}{3}$, the change in kinetic energy is $\Delta K = qdE \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2} qEd$. Thus, option (C) is incorrect. For option (D), after $t \gt t_f$, the electric field is turned off. The net torque on the dipole becomes zero. By Newton's first law of rotational motion, the dipole will continue to rotate about its center of mass with a constant angular velocity. Thus, option (D) is correct. Answer: If the magnitude of the final angular velocity $\omega_f = \sqrt{\dfrac{2qE}{md}}$, then $\theta_f = \dfrac{\pi}{6}$.; For $\theta_f = \pi/4$, the dipole rotates around its center of mass with a constant angular velocity after $t \gt t_f$.
2. Q2JEE Advanced Adv 2026 (Paper 2)
   Two charges $Q_1 = q$ and $Q_2 = mq$ are placed at the points $P_1(a, b)$ and $P_2(ma, mb)$, respectively, in the $XY$ plane, where $a, b \neq 0$ and $m \neq 0, 1$. If $V_1$ is the potential at a point in the $XY$ plane due to charge $Q_1$ and $V_2$ is the potential at that point due to charge $Q_2$. Correct statement(s) for the points at which $|V_1| = |V_2|$ is/are:
   1. A.For $m = -1$, locus of these points is $ax + by = 0$.
   2. B.For $m = 2$, the locus of these points is a circle of radius $\dfrac{2}{3}\sqrt{a^2 + b^2}$ centered at $\left(\dfrac{2}{3}a, \dfrac{2}{3}b\right)$
   3. C.For $m = -2$, the locus of these points is a circle of radius $2\sqrt{a^2 + b^2}$ centered at $(2a, 2b)$
   4. D.For $m = -3$, locus of these points is $3bx + 3ay = 0$.
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   Answer: A,B,C

   The potential at a point $(x, y)$ due to charge $Q_1$ is $V_1 = \dfrac{1}{4\pi\epsilon_0} \dfrac{q}{\sqrt{(x-a)^2 + (y-b)^2}}$. The potential at $(x, y)$ due to charge $Q_2$ is $V_2 = \dfrac{1}{4\pi\epsilon_0} \dfrac{mq}{\sqrt{(x-ma)^2 + (y-mb)^2}}$. Given $|V_1| = |V_2|$, we have: $\dfrac{1}{(x-a)^2 + (y-b)^2} = \dfrac{m^2}{(x-ma)^2 + (y-mb)^2}$ Cross-multiplying and expanding both sides: $(x-ma)^2 + (y-mb)^2 = m^2[(x-a)^2 + (y-b)^2]$ $x^2 + m^2a^2 - 2max + y^2 + m^2b^2 - 2mby = m^2(x^2 + a^2 - 2ax + y^2 + b^2 - 2by)$ $x^2 + y^2 - 2m(ax+by) + m^2(a^2+b^2) = m^2(x^2+y^2) - 2m^2(ax+by) + m^2(a^2+b^2)$ Canceling $m^2(a^2+b^2)$ from both sides and rearranging: $(m^2-1)(x^2+y^2) - 2m(m-1)(ax+by) = 0$ Since $m \neq 1$, dividing by $m-1$ gives the general equation of the locus: $(m+1)(x^2+y^2) - 2m(ax+by) = 0$ For $m = -1$: $0 - 2(-1)(ax+by) = 0 \Rightarrow ax+by = 0$. This represents a straight line. For $m = 2$: $3(x^2+y^2) - 4(ax+by) = 0 \Rightarrow x^2+y^2 - \dfrac{4}{3}ax - \dfrac{4}{3}by = 0$. This represents a circle with center $\left(\dfrac{2}{3}a, \dfrac{2}{3}b\right)$ and radius $\sqrt{\left(\dfrac{2}{3}a\right)^2 + \left(\dfrac{2}{3}b\right)^2} = \dfrac{2}{3}\sqrt{a^2+b^2}$. For $m = -2$: $(-1)(x^2+y^2) - 2(-2)(ax+by) = 0 \Rightarrow x^2+y^2 - 4ax - 4by = 0$. This represents a circle with center $(2a, 2b)$ and radius $\sqrt{(2a)^2 + (2b)^2} = 2\sqrt{a^2+b^2}$. For $m = -3$: $(-2)(x^2+y^2) - 2(-3)(ax+by) = 0 \Rightarrow x^2+y^2 - 3ax - 3by = 0$. This represents a circle, not a straight line. Answer: For $m = -1$, locus of these points is $ax + by = 0$.; For $m = 2$, the locus of these points is a circle of radius $\dfrac{2}{3}\sqrt{a^2 + b^2}$ centered at $\left(\dfrac{2}{3}a, \dfrac{2}{3}b\right)$; For $m = -2$, the locus of these points is a circle of radius $2\sqrt{a^2 + b^2}$ centered at $(2a, 2b)$
3. Q3JEE Advanced Adv 2022 (Paper 2)
   A disk of radius $R$ with uniform positive charge density $\sigma$ is placed on the $xy$ plane with its center at the origin. The Coulomb potential along the $z$-axis is $V\left(z\right)=\dfrac{\sigma }{2{ϵ}_{0}}\left(\sqrt{{R}^{2}+{z}^{2}}-z\right)$ A particle of positive charge $q$ is placed initially at rest at a point on the $z$-axis with $z={z}_{0}$ and ${z}_{0}\gt 0$. In addition to the Coulomb force, the particle experiences a vertical force $\vec{F}=-c\hat{k}$ with $c\gt 0$. Let $\beta =\dfrac{2c{\epsilon }_{0}}{q\sigma }$. Which of the following statement(s) is(are) correct?
   1. A.For $\beta =\dfrac{1}{4}$ and ${z}_{0}=\dfrac{25}{7}R$, the particle reaches the origin.
   2. B.For $\beta =\dfrac{1}{4}$ and ${z}_{0}=\dfrac{3}{7}R$, the particle reaches the origin.
   3. C.For $\beta =\dfrac{1}{4}$ and ${z}_{0}=\dfrac{R}{\sqrt{3}}$, the particle returns back to $z={z}_{0}$
   4. D.For $\beta \gt 1$ and ${z}_{0}\gt 0$, the particle always reaches the origin.
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   Answer: A,C,D

   For the particle to reach at origin, work done by the force should be greater than the increase in the potential energy of the charged particle. Therefore, $cz\geq \dfrac{\sigma q}{2{ϵ}_{0}}\left[R-\left(\sqrt{{R}^{2}+{z}^{2}}-z\right)\right] \Rightarrow \dfrac{c2{ϵ}_{0}}{\sigma q}\geq \dfrac{\left[R-\left(\sqrt{{R}^{2}+{z}^{2}}-z\right)\right]}{z} \Rightarrow \beta \geq \dfrac{\left[R-\left(\sqrt{{R}^{2}+{z}^{2}}-z\right)\right]}{z}$ For option A $\dfrac{\left[R-\left(\sqrt{{R}^{2}+{z}^{2}}-z\right)\right]}{z}=\dfrac{R-R\left(\sqrt{1+\dfrac{625}{49}}-\dfrac{25}{7}\right)}{R\dfrac{25}{7}}≃0.242$ Since, $0.25\gt 0.242$ Option A is correct. For option B $\dfrac{\left[R-\left(\sqrt{{R}^{2}+{z}^{2}}-z\right)\right]}{z}=\dfrac{R-R\left(\sqrt{1+\dfrac{9}{49}}-\dfrac{3}{7}\right)}{R\dfrac{3}{7}}≃0.79$ Since, $0.25\lt 0.79$ Option B is incorrect For option C $\dfrac{\left[R-\left(\sqrt{{R}^{2}+{z}^{2}}-z\right)\right]}{z}=\dfrac{R-R\left(\sqrt{1+\dfrac{1}{3}}-\dfrac{1}{\sqrt{3}}\right)}{R\dfrac{1}{\sqrt{3}}}≃0.73$ Since, $0.25\lt 0.73$ Therefore, the particle returns to ${z}_{0}$ Hence, option C is correct. For option D, For any value of $z\gt 0$ $\dfrac{\left[R-\left(\sqrt{{R}^{2}+{z}^{2}}-z\right)\right]}{z}\lt 1$ Hence, option D is correct.
4. Q4JEE Advanced Adv 2020 (Paper 1)
   A circular disc of radius $R$ carries surface charge density $\sigma (r)={\sigma }_{0}\left(1-\dfrac{r}{R}\right)$, where ${\sigma }_{0}$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is ${\phi }_{0}$. Electric flux through another spherical surface of radius $\dfrac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\dfrac{{\phi }_{0}}{\phi }$ is
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   Answer: 6.4

   Consider a ring element of radius $r$, thickness in disc charge of element $dQ=\left(2\pi rdr\right)\sigma \Rightarrow dQ=2\pi r\sigma \left(dr\right)$ Total charge, $Q=\int 2\pi r{\sigma }_{0}\left(1-\dfrac{r}{R}\right)dr$ $Q=2\pi {\sigma }_{0}\left(\dfrac{{r}^{2}}{2}-\dfrac{{r}^{3}}{3R}\right)$ ${\phi }_{0}=\dfrac{\text{charge upto}R}{{ϵ}_{0}}=2\pi {\sigma }_{0}\left(\dfrac{{R}^{2}}{2}-\dfrac{{R}^{3}}{3R}\right)$ $\phi =\dfrac{\text{charge upto}\dfrac{R}{4}}{{ϵ}_{0}}=2\pi {\sigma }_{0}\left(\dfrac{{R}^{2}}{16\times 2}-\dfrac{{R}^{3}}{64\times 3R}\right)$ $\dfrac{{\phi }_{0}}{\phi }=\dfrac{\dfrac{1}{2}-\dfrac{1}{3}}{\dfrac{1}{32}-\dfrac{1}{192}}=\dfrac{1}{6}\times \dfrac{32\times 6}{5}=6.4$
5. Q5JEE Advanced Adv 2014 (Paper 1)
   Let ${E}_{1}\left(r\right),{E}_{2}\left(r\right)$ and ${E}_{3}\left(r\right)$ be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density $\lambda$ , and an infinite plane with uniform surface charge density $\sigma$ . If ${E}_{1}\left({r}_{0}\right)={E}_{2}\left({r}_{0}\right)={E}_{3}\left({r}_{0}\right)$ at a given distance ${r}_{0}$ , then
   1. A.$Q=4\sigma \pi {r}_{0}^{2}$
   2. B.${r}_{0}=\dfrac{\lambda }{2\pi \sigma }$
   3. C.${E}_{1}\left(\dfrac{{r}_{0}}{2}\right)=2{E}_{2}\left(\dfrac{{r}_{0}}{2}\right)$
   4. D.${E}_{2}\left(\dfrac{{r}_{0}}{2}\right)=4{E}_{3}\left(\dfrac{{r}_{0}}{2}\right)$
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   Answer: (C)

   $\dfrac{Q}{4\pi {\epsilon }_{0}{r}_{0}^{2}}=\dfrac{\lambda }{2\pi {\epsilon }_{0}{r}_{0}}=\dfrac{\sigma }{2{\epsilon }_{0}}$ ${E}_{1}\left(\dfrac{{r}_{0}}{2}\right)=\dfrac{Q}{\pi {\epsilon }_{0}{r}_{0}^{2}},{E}_{2}\left(\dfrac{{r}_{0}}{2}\right)=\dfrac{\lambda }{\pi {\epsilon }_{0}{r}_{0}},{E}_{3}\left(\dfrac{{r}_{0}}{2}\right)=\dfrac{\sigma }{2{\epsilon }_{0}}$ ${E}_{1}\left(\dfrac{{r}_{0}}{2}\right)=2{E}_{2}\left(\dfrac{{r}_{0}}{2}\right)$