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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:
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:
Solution:
1528 Upvotes
Verified Answer
The correct answer is:
A, C and E only
Given $V=V=$ Constant
(i)
$\begin{aligned}
& C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d} \\
& d^{\prime \prime} \lt d \\
& C^{\prime}\gtC
\end{aligned}$
Hence, final capacitance greater than initial capacitance,
(ii)
$\begin{aligned}
U^{\prime} & =\frac{1}{2} C^{\prime} V^2 \\
U & =\frac{1}{2} C V^2 \\
U^{\prime \prime} & \gtU
\end{aligned}$
Hence final energy is greater than initial energy
(iii)
$\begin{aligned}
& \frac{Q^{\prime}}{V^{\prime}}=C^{\prime} \text { and } \frac{Q}{V}=C \\
& \frac{Q^{\prime}}{V^{\prime}} \neq \frac{Q}{V}
\end{aligned}$
(iv) Product of charge and voltage
$\begin{aligned}
& X^{\prime}=Q^{\prime} V=C^{\prime} V^2 \\
& X=Q V=C V^2 \\
& X^{\prime}\gtX
\end{aligned}$
(i)
$\begin{aligned}
& C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d} \\
& d^{\prime \prime} \lt d \\
& C^{\prime}\gtC
\end{aligned}$
Hence, final capacitance greater than initial capacitance,
(ii)
$\begin{aligned}
U^{\prime} & =\frac{1}{2} C^{\prime} V^2 \\
U & =\frac{1}{2} C V^2 \\
U^{\prime \prime} & \gtU
\end{aligned}$
Hence final energy is greater than initial energy
(iii)
$\begin{aligned}
& \frac{Q^{\prime}}{V^{\prime}}=C^{\prime} \text { and } \frac{Q}{V}=C \\
& \frac{Q^{\prime}}{V^{\prime}} \neq \frac{Q}{V}
\end{aligned}$
(iv) Product of charge and voltage
$\begin{aligned}
& X^{\prime}=Q^{\prime} V=C^{\prime} V^2 \\
& X=Q V=C V^2 \\
& X^{\prime}\gtX
\end{aligned}$
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