Electrostatic JEE Main PYQs (2020–2026)

Solved Electrostatic PYQs

3 recent questions with answers. The full set of 206 is free to practise.

1. JEE Main 2026EasyElectrostatic Potential and Equipotential Surfaces
   Q1

   The electric potential as a function of $x,y$ is given by $=5\left(x^2-y^2\right)V$ . The electric field at a point $(2,3)m$ is $\_\_\_\_V/m$ .

   • A$(-20\hat{i}+30\hat{j})$
   • B$(20\hat{i}-30\hat{j})$
   • C$(20\hat{i}+45\hat{j})$
   • D$(-4\hat{i}+6\hat{j})$

   Show answer & solution

   Correct answer: (A)

   The electric field $\overrightarrow{E}$ is the negative gradient of the electric potential $V$ :
   
   $$\overrightarrow{E}=-\nabla V=-\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)$$
   
   The electric potential is given as $=5\left(x^2-y^2\right)=5x^2-5y^2$ .
   
   The x -component of electric field is,
   
   $$E_x=-\frac{\partial V}{\partial x}=-\frac{\partial }{\partial x}\left(5x^2-5y^2\right)=-(10x)=-10x$$
   
   The y -component of electric field is,
   
   $$E_y=-\frac{\partial V}{\partial y}=-\frac{\partial }{\partial y}\left(5x^2-5y^2\right)=(10y)=10x$$
   
   At Point $(2,3)$
   
   $$E_x=-10(2)=-20V/m$$
   
   $$\Rightarrow E_y=10(3)=30V/m$$
   
   So, the resulting electric field vector is :
   
   $$\overrightarrow{E}=-20\hat{ı}+30\hat{ȷ}$$
   
   Therefore, the correct option is (A).
2. JEE Main 2026MediumElectric Field and Motion of Charged Particles
   Q2

   A thin half ring of radius 35 cm is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100V/m$ , then the value of $Q$ is $\_\_\_\_$ nC .
   
   $$\left(\begin{array}{r}{ϵ}_0=8.85\times \frac{10^{-12}{C}^2}{{Nm}^2}\\ \text{ and }\pi =3.14\end{array}\right)$$

   • A2.14
   • B2.44
   • C3.25
   • D0.7

   Show answer & solution

   Correct answer: (A)

   For a uniformly charged semicircular ring with linear charge density $\lambda$ and radius R , the electric field E at the center is:
   
   $$E=\frac{2k\lambda }{R}=\frac{2\lambda }{4\pi {ϵ}_{0}R}=\frac{\lambda }{2\pi {ϵ}_{0}R}$$
   
   Where $\lambda =\frac{\text{ Total Charge }}{\text{ Length }}=\frac{Q}{\pi R}$ .
   
   Substituting $\lambda$ into the field formula:
   
   $$E=\frac{Q/\pi R}{2\pi {ϵ}_{0}R}=\frac{Q}{2{\pi }^2{ϵ}_0{R}^2}$$
   
   $$\Rightarrow Q=E\cdot 2{\pi }^2{ϵ}_0{R}^2$$
   
   The magnitude of electric field at the centre is $E=100V/m$
   
   The radius of the semi-circle is $R=35cm=0.35m$
   
   The value of ${ϵ}_0=8.85\times 10^{-12}$ and $\pi =3.14$
   
   $$Q=100\times 2\times (3.14{)}^2\times 8.85\times 10^{-12}\times (0.35{)}^2$$
   
   $$\Rightarrow Q=100\times 2\times 9.8596\times 8.85\times 10^{-12}\times 0.1225$$
   
   $$\Rightarrow Q\approx 2.14\times 10^{-9}C$$
   
   $$\Rightarrow Q=2.14nC$$
   
   Therefore, the value of Q is 2.14 nC . Thus, the correct option is (A).
3. JEE Main 2026MediumElectrostatic Potential Energy and Work Done
   Q3

   Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm . The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is $\_\_\_\_\mu J$.
   
   $$\frac{1}{4\pi {ϵ}_0}=9\times 10^{9}N\cdot m^2/C^2$$

   • A3.3
   • B27
   • C2.7
   • D5.4

   Show answer & solution

   Correct answer: (C)

   When 3 nC is at infinity then electric potential due to charges 1 nC and 2 nC is zero. So the initial potential is $V_i$
   
   The total potential at the third corner due to $q_1(1nC)$ and $q_2(2nC)$ is the sum of their individual potentials:
   
   $$V_f=\frac{{kq}_1}{r}+\frac{{kq}_2}{r}=\frac{k\left(q_1+q_2\right)}{r}$$
   
   The work done is given by $W=q_3\times \Delta V$ , where $\Delta V=$ potential difference between initial and final position.
   
   $$W=q_3\times \left[\frac{k\left(q_1+q_2\right)}{r}-0\right]$$
   
   $$\Rightarrow W=\frac{k\times q_3\times \left(q_1+q_2\right)}{r}$$
   
   Substituting the values,
   
   $$W=\frac{\left(9\times 10^9\right)\times \left(3\times 10^{-9}\right)\times \left(1\times 10^{-9}+2\times 10^{-9}\right)}{0.03}$$
   
   $$\Rightarrow W=\frac{(27)\times \left(3\times 10^{-9}\right)}{0.03}=27\times 10^{-7}J$$
   
   As $1J=10^6\mu J$
   
   $$W=27\times 10^{-7}\times 10^6\mu J$$
   
   $$\Rightarrow W=27\times 10^{-1}\mu J=2.7\mu J$$
   
   So, the work done in bring a charge of 3 nC from infinity to the third corner of the triangle is $2.7\mu J$ .
   
   Hence, the correct option is (C).

Topic-wise question count

Where JEE Main has asked from Electrostatic, most frequent first.

• Electric Field and Motion of Charged Particles54
• Electric Flux and Gauss’s Law38
• Electric Charge and Coulomb's Law33
• Electric Dipole27
• Conductors in Electrostatic Equilibrium and Spherical Shells20
• Electrostatic Potential and Equipotential Surfaces18
• Electrostatic Potential Energy and Work Done16

Year-wise Electrostatic questions in JEE Main

Number of questions from this chapter in each year's paper(s).