Capacitance JEE Main previous year questions with solutions

5 solved JEE Main questions on Capacitance, free to read — no sign-in needed. The full chapter has 176 questions; sign in to attempt the remaining 171 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (04 Apr, Shift 1)Capacitance
   A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed $v$. 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 _____.
   1. A.$-2$
   2. B.$1$
   3. C.$-1$
   4. D.$2$
   Show answer & solution

   Answer: (A)

   The electrostatic energy of a parallel plate capacitor is given by $U = \dfrac{1}{2} C V^2$. For a parallel plate capacitor, the capacitance is $C = \dfrac{\varepsilon_0 A}{x}$, where $x$ is the separation between the plates. Since the capacitor remains connected to the battery, the potential difference $V$ across the plates is constant. Substituting $C$ into the energy equation gives: $U = \dfrac{\varepsilon_0 A V^2}{2x}$ Differentiating $U$ with respect to time $t$ to find the time rate of change of electrostatic energy: $\dfrac{dU}{dt} = \dfrac{d}{dt} \left( \dfrac{\varepsilon_0 A V^2}{2x} \right) = -\dfrac{\varepsilon_0 A V^2}{2x^2} \dfrac{dx}{dt}$ Given that the plates are pulled apart at a uniform speed $v$, we have $\dfrac{dx}{dt} = v$. $\dfrac{dU}{dt} = -\dfrac{\varepsilon_0 A V^2 v}{2x^2}$ From the above expression, it is clear that $\dfrac{dU}{dt} \propto x^{-2}$. Comparing this with $x^\alpha$, we get $\alpha = -2$. Answer: $-2$
2. Q2JEE Main 2025 (07 Apr, Shift 2)Capacitors with Dielectric
   A parallel plate capacitor has charge $5 \times 10^{-6} \mathrm{C}$. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 10^{-6} \mathrm{C}$ then the dielectric constant of the slab is ________.
   Show answer & solution

   Answer: 5

   $\begin{aligned} & \mathrm{Q}_{\mathrm{in}}=\mathrm{Q}\left(1-\frac{1}{\mathrm{~K}}\right) \\ & 4 \times 10^{-6}=5 \times 10^{-6}\left(1-\frac{1}{\mathrm{~K}}\right) \\ & 1-\frac{1}{\mathrm{~K}}=\frac{4}{5} \\ & \mathrm{~K}=5\end{aligned}$
3. Q3JEE Main 2024 (05 Apr, Shift 1)Charging and Discarging of capacitors
   The electric field between the two parallel plates of a capacitor of $1.5 \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is _____$\Omega$. (Given, $\log 3=1.1$ )
   Show answer & solution

   Answer: 4

   $\begin{aligned} & \mathrm{E}=\frac{\mathrm{E}_0}{3} \Rightarrow \mathrm{V}=\frac{\mathrm{V}_0}{3} \\ & \frac{\mathrm{V}_0}{3}=\mathrm{V}_0 \mathrm{e}^{-\frac{\mathrm{t}}{\tau}} \\ & \mathrm{t}=\tau \ell \mathrm{n} 3 \\ & 6.6 \times 10^{-6}=\mathrm{R}\left(1.5 \times 10^{-6}\right)(1.1) \\ & \mathrm{R}=\frac{6}{1.5}=4 \Omega\end{aligned}$
4. Q4JEE Main 2022 (29 Jun, Shift 1)Force and Energy stored in capacitor
   A parallel plate capacitor filled with a medium of dielectric constant $10$, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$. Then the energy of capacitor will
   1. A.increase by $50$
   2. B.decrease by $15$
   3. C.increase by $25$
   4. D.increase by $33$
   Show answer & solution

   Answer: (A)

   When a dielectric of dielectric constant $K$ is inserted between the plates of a capacitor ${C}_{0}$, the capacitance becomes $K{C}_{0}$. Energy stored in a capacitor is given by, $U=\dfrac{1}{2}C{V}^{2}$. Now, ${U}_{i}=\dfrac{1}{2}\left({K}_{1}{C}_{0}\right){V}^{2}$ and ${U}_{f}=\dfrac{1}{2}\left({K}_{2}{C}_{0}\right){V}^{2}$ $\Rightarrow \Delta U={U}_{f}-{U}_{i}=\dfrac{1}{2}\left({K}_{2}-{K}_{1}\right){C}_{0}{V}^{2}$ Percentage change in the energy will be, $\dfrac{\Delta U}{{U}_{i}}\times 100=\dfrac{\dfrac{1}{2}\times 5\times {{C}_{0}}^{2}}{\dfrac{1}{2}\times 10\times {{C}_{0}}^{2}}\times 100=50$
5. Q5JEE Main 2021 (26 Feb, Shift 1)Grouping of capacitors
   Consider the combination of two capacitors ${C}_{1}$ and ${C}_{2}$, with ${C}_{2}\gt {C}_{1}$, when connected in parallel, the equivalent capacitance is $10$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $\dfrac{{C}_{2}}{{C}_{1}}$.
   1. A.$4+\sqrt{15}$
   2. B.$2+\sqrt{15}$
   3. C.$9$
   4. D.$\dfrac{15}{4}$
   Show answer & solution

   Answer: (A)

   When connected in parallel ${C}_{eq}={C}_{1}+{C}_{2}$ When in series ${C}_{eq}^{'}=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}$ ${C}_{1}+{C}_{2}=10\left(\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}\right)$ ${\left({C}_{1}+{C}_{2}\right)}^{2}=10{C}_{1}{C}_{2}$ ${C}_{1}^{2}+{C}_{2}^{2}-8{C}_{1}{C}_{2}=0$ dividing by ${C}_{1}^{2}$ ${\left(\dfrac{{C}_{2}}{{C}_{1}}\right)}^{2}-\dfrac{8{C}_{2}}{{C}_{1}}+1=0$ Let $\dfrac{{C}_{2}}{{C}_{1}}=x$ ${x}^{2}-8x+1=0$ $x=\dfrac{8\pm \sqrt{64-4}}{2}=4+\sqrt{15}$ This question was Bonus in JEE Main 2021 exam because of incorrect data in original question.