Question 611: Electrostatics & Capacitance

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:

Accepted Answer

Given $$V^{\prime}=V=$$ Constant
$$\begin{aligned}
	ext{(i)}\quad & C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d} \
& d^{\prime} C
\end{aligned}$$
Hence, final capacitance greater than initial capacitance,
$$
\begin{aligned}
	ext { (ii) } \quad U^{\prime} & =\frac{1}{2} C^{\prime} V^2 \
U & =\frac{1}{2} C V^2 \
U^{\prime} & >U
\end{aligned}
$$
Hence final energy is greater than initial energy
$$
\begin{aligned}
	ext { (iii) }\quad \frac{Q^{\prime}}{V^{\prime}} & =C^{\prime} 	ext { and } \frac{Q}{V}=C \
\frac{Q^{\prime}}{V^{\prime}} & 
eq \frac{Q}{V}
\end{aligned}$$
(iv) Product of charge and voltage
$$\begin{aligned}
& X^{\prime}=Q^{\prime} V=C^{\prime} V^2 \
& X=Q V=C V^2 \
& X^{\prime}>X
\end{aligned}$$

Answer

A, B and E only

Answer

A, C and E only

Answer

B, D and E only

Answer

A, B and C only

Electrostatics & Capacitance

Choose the correct answer: