પ્લેટનું ક્ષેત્રફળ A અને બે પ્લેટ વચ્ચેનું અં | vedclass

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Question

(b) After inserting the dielectric slab
New capacitance $C' = K.C = \frac{{K{\varepsilon _0}A}}{d}$
New potential difference $V' = \frac{V}{

Vedclass · Accepted Answer

$W = \frac{{{\varepsilon _0}A{V^2}}}{{2kd}}$

Vedclass · Answer

(b) After inserting the dielectric slab
New capacitance $C' = K.C = \frac{{K{\varepsilon _0}A}}{d}$
New potential difference $V' = \frac{V}{K}$
New charge $Q' = C'V' = \frac{{{\varepsilon _0}AV}}{d}$
New electric field $E' = \frac{{V'}}{d} = \frac{V}{{Kd}}$
Work done $(W) =$ Final energy $-$ Initial energy
W $ = \frac{1}{2}C'V{'^{\,2}} - \frac{1}{2}C{V^2}$ $ = \frac{1}{2}(KC)\,{\left( {\frac{V}{K}} \right)^2} - \frac{1}{2}C{V^2}$
$ = \frac{1}{2}C{V^2}\left( {\frac{1}{K} - 1} \right) = - \frac{1}{2}C{V^2}\left( {1 - \frac{1}{K}} \right)$
$ = - \frac{{{\varepsilon _0}A{V^2}}}{{2d}}\left( {1 - \frac{1}{K}} \right)$ so $|W|\, = \frac{{{\varepsilon _0}A{V^2}}}{{2d}}\left( {1 - \frac{1}{K}} \right)$.

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