If the plates of a parallel plate capacitor c | vedclass

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Question

(A) Given that the capacitor is connected to a battery,the potential difference $V$ remains constant.$(i)$ Capacitance $C = \frac{\varepsilon_0 A}{d}$

Vedclass · Accepted Answer

$A, C$ and $E$ only

Vedclass · Answer

(A) Given that the capacitor is connected to a battery,the potential difference $V$ remains constant.$(i)$ Capacitance $C = \frac{\varepsilon_0 A}{d}$. As the plates are moved closer,the distance $d$ decreases,so the capacitance $C$ increases. Thus,statement $C$ is correct.$(ii)$ Charge $Q = CV$. Since $C$ increases and $V$ is constant,the charge $Q$ increases. Thus,statement $A$ is correct.$(iii)$ Energy stored $U = \frac{1}{2}CV^2$. Since $C$ increases and $V$ is constant,the energy $U$ increases. Thus,statement $B$ is incorrect.$(iv)$ The ratio of charge to potential is $\frac{Q}{V} = C$. Since $C$ increases,the ratio $\frac{Q}{V}$ changes. Thus,statement $D$ is incorrect.$(v)$ The product of charge and voltage is $QV = CV^2$. Since $C$ increases and $V$ is constant,the product $QV$ increases. Thus,statement $E$ is correct.Therefore,statements $A, C,$ and $E$ are correct.

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