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Question

(B) In Figure $(a)$,the capacitor is filled with a dielectric of thickness $d$. The capacitance is $C = \frac{K \varepsilon_0 A}{d}$.In Figure $(b)$,t

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The capacitance is decreased by a factor of $\frac{1}{K+1}$.

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(B) In Figure $(a)$,the capacitor is filled with a dielectric of thickness $d$. The capacitance is $C = \frac{K \varepsilon_0 A}{d}$.In Figure $(b)$,the total distance between the plates is $2d$. The dielectric slab of thickness $d$ is between the plates,leaving a vacuum gap of $d$ (i.e.,$d/2$ on each side). The system acts as two capacitors in series: one with dielectric (thickness $d$,capacitance $C_1 = \frac{K \varepsilon_0 A}{d}$) and one with air (thickness $d$,capacitance $C_2 = \frac{\varepsilon_0 A}{d}$).The equivalent capacitance $C'$ is given by $\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{K \varepsilon_0 A} + \frac{d}{\varepsilon_0 A} = \frac{d}{\varepsilon_0 A} (\frac{1}{K} + 1) = \frac{d(K+1)}{K \varepsilon_0 A}$.Thus,$C' = \frac{K \varepsilon_0 A}{d(K+1)} = \frac{C}{K+1}$.Therefore,the capacitance decreases by a factor of $(K+1)$,which means it is multiplied by $\frac{1}{K+1}$. Hence,option $(B)$ is correct.

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