Question
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
Options
Solution
Sol. [Key Concept: Capacitor plates moved apart while connected to a battery]
Given $V^{'} = V =$ Constant
(i) $C^{'} = \frac{\varepsilon_{0}A}{d^{'}},C = \frac{\varepsilon_{0}A}{d}$
$$\begin{matrix} & d^{'} < d \\ & C^{'} > C \end{matrix}$$
Hence, final capacitance greater than initial capacitance,
(ii) $U^{'} = \frac{1}{2}C^{'}V^{2}$
$$\begin{matrix} & U = \frac{1}{2}CV^{2} \\ & U^{'} > U \end{matrix}$$
Hence final energy is greater than initial energy
(iii) $\frac{Q^{'}}{V^{'}} = C^{'}$ and $\frac{Q}{V} = C$
$$\frac{Q^{'}}{V^{'}} \neq \frac{Q}{V}$$
(iv) Product of charge and voltage
$$\begin{matrix} & X^{'} = Q^{'}V = C^{'}V^{2} \\ & X = QV = CV^{2} \\ & X^{'} > X \end{matrix}$$
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