JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 21)
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
240
Explanation
$${d_1} = 4 \times {10^{ - 3}}$$
$${A_1} = 8 \times 4 \times {10^{ - 4}}\,{m^2}$$
$$V = 20\,V$$
$${d_2} = 4 \times {10^{ - 3}},$$
$${A_2} = 4 \times 1 \times {10^{ - 4}}\,{m^2}$$
$${C_{eq}} = {{({A_1} + 5{A_2} - {A_2}){\varepsilon _0}} \over d} = {{3(16) \times {{10}^{ - 4}}} \over {4 \times {{10}^{ - 3}}}}{\varepsilon _0}$$
$$\varepsilon = {1 \over 2}{C_{eq}}{V^2} = {3 \over 2}\left( {{4 \over {10}}} \right)(400){\varepsilon _0} = 240{\varepsilon _0}$$
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