JEE Main 2024 — Capacitance Question with Solution
JEE Main 2024 (06 Apr Shift 2)
Question
A capacitor of capacitance whose plates are separated by through air and each plate has area is now filled equally with two dielectric media of respectively as shown in figure. If new force between the plates is . The supply voltage is _________ .
(we modified language of question to make it correct)
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Show full solutionCorrect answer: 93
Correct answer
93
Step-by-step explanation
$\begin{aligned} & \mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2 \\ & \mathrm{C}_1=\frac{2 \epsilon_0 \mathrm{~A}}{2 \times \mathrm{d}}=10 \mu \mathrm{F} \\ & \mathrm{C}_2=\frac{3 \epsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=15 \mu \mathrm{F} \\ & \mathrm{C}_{\mathrm{eq}}=25 \mu \mathrm{F} \end{aligned}$
Now the charge on $\begin{aligned} & \mathrm{C}_1=10 \mathrm{~V} \mu \mathrm{c} \\ & \mathrm{C}_2=1.5 \mathrm{~V} \mu \mathrm{C} . \end{aligned}\left[F=\frac{Q^2}{2 A \in_0}\right]$ $\begin{aligned} & \frac{100 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \epsilon_0}+\frac{225 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \epsilon_0}=8 \\ & 325 \mathrm{~V}^2=8 \times 4 \times 10^{-4} \times 8.85 \\ & \mathrm{~V}^2=\frac{32 \times 8.85 \times 10^{-4}}{325} \\ & \therefore \mathrm{V}=\sqrt{\frac{283.2 \times 10^{-4}}{325}} \\ & \mathrm{~V}=0.93 \times 10^{-2} \end{aligned}$
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This is a previous-year question from JEE Main 2024, covering the Capacitance chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.