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A parallel plate capacitor with width $4 \mathrm{~cm}$ , length $8 \mathrm{~cm}$ and separation between the plates of $4 \mathrm{~mm}$ is connected to a battery of $20 \mathrm{~V}$ . A dielectric slab of dielectric constant 5 having length $1 \mathrm{~cm}$ , width $4 \mathrm{~cm}$ and thickness $4 \mathrm{~mm}$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be___________ $\mathrm{\epsilon_{0} J}$ . (Where $\mathrm{\epsilon_{0} }$ is the permittivity of free space)
Option: 1
240

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{Area=length\times width=32\times 10^{-4}m^{2}}$

$\mathrm{C_{eq}=C_{1}+C_{2}}$

$\mathrm{C_{eq}=\frac{7A\varepsilon _{0}}{8d}+\frac{A\varepsilon _{0}k}{8\left ( d \right )}}$

$\mathrm{=\frac{A\varepsilon _{0}}{d}\left ( \frac{7}{8}+\frac{5}{8} \right )}$

$\mathrm{C_{eq}=\frac{3A\varepsilon _{0}}{2d}}$

Eletrostatic energy of the system is

$\mathrm{U=\frac{1}{2}C_{eq}V^{2}}$

$\mathrm{\frac{3}{4}\times \frac{32\times 10^{-4}\varepsilon _{0}}{4\times 10^{-3}}\times 400}$

$\mathrm{U=240\varepsilon _{0}J}$

The electrostatic energy of this system will be $\mathrm{240\varepsilon _{0}J}$

Posted by

Kuldeep Maurya

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