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 ____________ $\epsilon_{0}$ J. (Where $\epsilon_{0}$ is the permittivity of free space)
Answer (integer)
240
Solution
<p>${d_1} = 4 \times {10^{ - 3}}$</p>
<p>${A_1} = 8 \times 4 \times {10^{ - 4}}\,{m^2}$</p>
<p>$V = 20\,V$</p>
<p>${d_2} = 4 \times {10^{ - 3}},$</p>
<p>${A_2} = 4 \times 1 \times {10^{ - 4}}\,{m^2}$</p>
<p>$${C_{eq}} = {{({A_1} + 5{A_2} - {A_2}){\varepsilon _0}} \over d} = {{3(16) \times {{10}^{ - 4}}} \over {4 \times {{10}^{ - 3}}}}{\varepsilon _0}$$</p>
<p>$$\varepsilon = {1 \over 2}{C_{eq}}{V^2} = {3 \over 2}\left( {{4 \over {10}}} \right)(400){\varepsilon _0} = 240{\varepsilon _0}$$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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