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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2024 · 4 April · Evening Shift · Physics · Capacitance

A parallel plate capacitor of capacitance 12.5 pF is charged by a...

A parallel plate capacitor of capacitance 12.5pF12.5 \mathrm{pF} is charged by a battery connected between its plates to potential difference of 12.0 V12.0 \mathrm{~V}. The battery is now disconnected and a dielectric slab (ϵr=6)\left(\epsilon_{\mathrm{r}}=6\right) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _____ 10−12 J10^{-12} \mathrm{~J}.

Correct answer

0

Step-by-step solution

Before inserting dielectric capacitance is given C0=12.5pF\mathrm{C}_0=12.5 \mathrm{pF} and charge on the capacitor Q=C0 V\mathrm{Q}=\mathrm{C}_0 \mathrm{~V} After inserting dielectric capacitance will become ϵrC0\epsilon_{\mathrm{r}} \mathrm{C}_0. Change in potential energy of the capacitor =Ei−Ef=Q22Ci−Q22Cf=Q22C0[1−1ϵr]=(C0 V)22C0[1−1ϵr]=12C0 V2[1−1ϵr]\begin{aligned} & =\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{f}} \\ & =\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right] \\ & =\frac{\left(\mathrm{C}_0 \mathrm{~V}\right)^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_0 \mathrm{~V}^2\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right] \end{aligned}
Using C0=12.5pF,V=12 V,ϵr=6\mathrm{C}_0=12.5 \mathrm{pF}, \mathrm{V}=12 \mathrm{~V}, \epsilon_{\mathrm{r}}=6=12(12.5)×122[1−16]=12(12.5)×122×56=750pJ=750×10−12 J\begin{aligned} & =\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^2 \times \frac{5}{6} \\ & =750 \mathrm{pJ}=750 \times 10^{-12} \mathrm{~J} \end{aligned}$$
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