Capacitor JEE Main PYQs (2020–2026)

Solved Capacitor PYQs

3 recent questions with answers. The full set of 122 is free to practise.

1. JEE Main 2026MediumDielectrics in Capacitors
   Q1

   A parallel plate capacitor is having separation between plates 0.885 mm . It has a capacitance of $1\mu F$ when the space between the plates is filled with an insulating material of resistivity $1\times 10^{13}\Omega m$ and resistance $17.7\times 10^{14}\Omega$ . Relative permittivity of the insulating material is $\times 10^7$ . The value of $\alpha$ is $\_\_\_\_$ .
   
   (Take permittivity of free space $=8.85\times 10^{-12}F/m$ )

   Numerical answer type

   Show answer & solution

   Correct answer: 2

   The resistance of a conductor of length d and cross-sectional area A is :
   
   $$R=\frac{\rho d}{A}$$
   
   Where, $\rho$ is the resistivity of the material.
   
   The capacitance of a parallel plate capacitor with plate separation d and area A is :
   
   $$C=\frac{ϵA}{d}=\frac{{ϵ}_r{ϵ}_{0}A}{d}$$
   
   $$\Rightarrow R\times C=\left(\frac{\rho d}{A}\right)\times \left(\frac{{ϵ}_r{ϵ}_{0}A}{d}\right)$$
   
   $$\Rightarrow RC=\rho {ϵ}_r{ϵ}_0$$
   
   $$\Rightarrow {ϵ}_r=\frac{RC}{\rho {ϵ}_0}$$
   
   The resistance of material is $R=17.7\times 10^{14}\Omega$ .
   
   The capacitance is $C=1\mu F=1\times 10^{-6}F$
   
   The resistivity is $\rho =1\times 10^{13}\Omega m$
   
   And the permittivity of free space is ${ϵ}_0=8.85\times 10^{-12}F/m$
   
   Putting the values for relative permittivity,
   
   $${ϵ}_r=\frac{\left(17.7\times 10^{14}\right)\times \left(1\times 10^{-6}\right)}{\left(1\times 10^{13}\right)\times \left(8.85\times 10^{-12}\right)}$$
   
   $$\Rightarrow {ϵ}_r=\frac{17.7\times 10^8}{8.85\times 10^1}$$
   
   $$\Rightarrow {ϵ}_r=2\times 10^7=\alpha \times 10^7\Rightarrow \alpha =2$$
   
   Therefore, the value of $\alpha$ is 2 .
2. JEE Main 2026MediumEnergy Stored in Capacitors and Redistribution of Charge
   Q2

   A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed . If $x$ is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to $x^{\alpha }$ , where $\alpha$ is $\_\_\_\_$ .

   • A-2
   • B1
   • C-1
   • D2

   Show answer & solution

   Correct answer: (A)

   The electrostatic energy $(U)$ stored in a capacitor connected to a battery of constant voltage $V$ is given by :
   
   $$U=\frac{1}{2}{CV}^2$$
   
   The capacitance $(C)$ of a parallel plate capacitor with plate area A and separation x is :
   
   $$C=\frac{{ϵ}_{0}A}{x}$$
   
   Substituting into the energy formula :
   
   $$U=\frac{1}{2}\left(\frac{{ϵ}_{0}A}{x}\right)V^2=\frac{{ϵ}_{0}A{V}^2}{2x}$$
   
   We need to find the time rate of change of energy, which is the derivative of $U$ with respect to time ( $t$ ) :
   
   $$\frac{dU}{dt}=\frac{d}{dt}\left(\frac{{ϵ}_0{AV}^2}{2x}\right)$$
   
   Since ${ϵ}_0,A$ , and $V$ are constants :
   
   $$\frac{dU}{dt}=\frac{{ϵ}_{0}A{V}^2}{2}\frac{d}{dt}\left(\frac{1}{x}\right)$$
   
   $$\Rightarrow \frac{dU}{dt}=\frac{{ϵ}_{0}A{V}^2}{2}\left(-\frac{1}{x^2}\right)\frac{dx}{dt}$$
   
   The plates are pulled apart at a uniform speed v , which means $\frac{dx}{dt}=v$ (a constant).
   
   $$\frac{dU}{dt}=-\left(\frac{{ϵ}_{0}A{V}^{2}v}{2}\right)\frac{1}{x^2}$$
   
   So, the rate of change of energy is proportional to $\frac{1}{x^2}$ :
   
   $$\frac{dU}{dt}\propto x^{-2}\Rightarrow x^{-2}=x^a\Rightarrow a=-2$$
   
   Therefore, the value of a is -2 .
   
   Thus, the correct option is (A).
3. JEE Main 2025MediumCapacitance and Parallel Plate Capacitor
   Q3

   A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A . If the rate of change of potential difference between the plates is $7\times 10^{8}V/s$ then the integer value of the distance between the parallel plates is
   
   $($ Take, ${ϵ}_0=9\times 10^{-12}\frac{F}{m},\pi =\frac{22}{7})$ ____________ $m$ .

   Numerical answer type

   Show answer & solution

   Correct answer: 1320

   Given, $r=10cm=\frac{1}{10}m$
   
   $I=0.15A$ , and $\frac{dv}{dt}=7\times 10^{8}v/s$
   
   We know, for a parallel plate capacitor,
   
   $c=\frac{{\epsilon }_{0}A}{d}\Rightarrow c=\frac{{\epsilon }_0\pi r^2}{d}$ …. (1)
   
   In a capacitor,
   
   $$Q=CV$$
   
   $$\Rightarrow V=\frac{Q}{C}$$
   
   by differentiating w.r.t. t
   
   $$\Rightarrow \frac{dv}{dt}=\frac{1}{C}\frac{dQ}{dt}$$
   
   $\Rightarrow \frac{dv}{dt}=\frac{d}{{\epsilon }_0\pi r^2}I$ (As $I=\frac{dQ}{dt}$ )
   
   $\Rightarrow d=\frac{{\epsilon }_0\pi r^2}{I}\frac{dv}{dt}$ …. (From (1))
   
   $$\Rightarrow d=\frac{9\times 10^{-12}}{0.15}\times \frac{22}{7}\times \frac{1}{100}\times 7\times 10$$
   
   $$=\frac{198}{0.15}\times 10^{-6}$$
   
   $=1320\times 10^{-6}$ m
   
   $$\Rightarrow d=1320\mu m$$

Topic-wise question count

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

• Dielectrics in Capacitors41
• Energy Stored in Capacitors and Redistribution of Charge23
• Capacitors in DC Circuits and RC Transients22
• Capacitance and Parallel Plate Capacitor20
• Combination of Capacitors and Equivalent Networks16

Year-wise Capacitor questions in JEE Main

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