A parallel plate capacitor having plate area | vedclass

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Question

(D) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$. Initial charge $q = CV = \frac{\varepsilon_0 A V}{d}$.Initial energy $U_i = \frac{q^2}{2C} =

Vedclass · Accepted Answer

$\frac{3 \varepsilon_0 A V^2}{2 d}$

Vedclass · Answer

(D) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$. Initial charge $q = CV = \frac{\varepsilon_0 A V}{d}$.Initial energy $U_i = \frac{q^2}{2C} = \frac{1}{2} CV^2 = \frac{\varepsilon_0 A V^2}{2d}$.When the battery is disconnected,the charge $q$ remains constant.The new separation is $d' = 4d$,so the new capacitance is $C' = \frac{\varepsilon_0 A}{4d} = \frac{C}{4}$.Final energy $U_f = \frac{q^2}{2C'} = \frac{q^2}{2(C/4)} = \frac{4q^2}{2C} = 4U_i$.Work done $W = U_f - U_i = 4U_i - U_i = 3U_i$.Substituting $U_i = \frac{\varepsilon_0 A V^2}{2d}$,we get $W = 3 	imes \frac{\varepsilon_0 A V^2}{2d} = \frac{3 \varepsilon_0 A V^2}{2d}$.

$A$ parallel plate capacitor having plate area $A$ and separation $d$ is charged to a potential difference $V$. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between plates is:

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