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 177 questions; sign in to attempt the remaining 172 in the exam simulator.

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 2023 (11 Apr, Shift 1)Force and Energy stored in capacitor
   The electric field in an electromagnetic wave is given as $\vec{E}=20\sin \omega \left(t-\dfrac{x}{c}\right)\vec{j}N{C}^{-1}$, where $\omega$ and $c$ are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of $5\times {10}^{-4}{m}^{3}$ will be (Given ${\epsilon }_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}{m}^{-2}$)
   1. A.$88.5\times {10}^{-13}J$
   2. B.$17.7\times {10}^{-13}J$
   3. C.$28.5\times {10}^{-13}J$
   4. D.$8.85\times {10}^{-13}J$
   Show answer & solution

   Answer: (D)

   Given that, $\vec{E}=20\sin \omega \left(t-\dfrac{x}{c}\right)\vec{j}N{C}^{-1}$ The average energy density of an electromagnetic wave is $\dfrac{1}{2}{\epsilon }_{0}{{E}^{2}}_{0}$. The total energy stored is ${E}_{\text{total}}=\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}\times volume$ $\Rightarrow {E}_{total}=\dfrac{1}{2}\times 8.85\times {10}^{-12}\times (20{)}^{2}\times 5\times {10}^{-4} =8.85\times {10}^{-13}J$
5. Q5JEE Main 2022 (26 Jul, Shift 1)Grouping of capacitors
   The total charge on the system of capacitance ${C}_{1}=1\mu F,{C}_{2}=2\mu F,{C}_{3}=4\mu F$ and ${C}_{4}=3\mu F$ connected in parallel is (Assume a battery of $20V$ is connected to the combination)
   1. A.$200\mu C$
   2. B.$200C$
   3. C.$10\mu C$
   4. D.$10C$
   Show answer & solution

   Answer: (A)

   As the given capacitors are in parallel combination, therefore,${C}_{eq}=\left(1+2+3+4\right)\mu F=10\mu F$ Hence, charge flown through the switch$∴Q=\left(20V\right)\times \left(10\mu F\right)=200\mu C$