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QJEEM22Ph5526: JEE Main 2022 Physics - Class 12 JEEM22Ph5526 Solution: A parallel plate capacitor with width , length and separation…

A parallel plate capacitor with width 4 cm4 \mathrm{~cm} , length 8 cm8 \mathrm{~cm} and separation between the plates of 4 mm4 \mathrm{~mm} is connected to a battery of 20 V20 \mathrm{~V} . A dielectric slab of dielectric constant 5 having length 1 cm1 \mathrm{~cm} , width 4 cm4 \mathrm{~cm} and thickness 4 mm4 \mathrm{~mm} is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ____________ ϵ0\epsilon_{0} J. (Where ϵ0\epsilon_{0} is the permittivity of free space)

Step-by-Step Verified Solution

d1=4×10−3{d_1} = 4 \times {10^{ - 3}}

A1=8×4×10−4 m2{A_1} = 8 \times 4 \times {10^{ - 4}}\,{m^2}

V=20 VV = 20\,V

d2=4×10−3,{d_2} = 4 \times {10^{ - 3}},

A2=4×1×10−4 m2{A_2} = 4 \times 1 \times {10^{ - 4}}\,{m^2}

Ceq=(A1+5A2−A2)ε0d=3(16)×10−44×10−3ε0{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}

ε=12CeqV2=32(410)(400)ε0=240ε0\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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