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If the distance between the plates of a parallel plate capacitor of capacity $10 \mu \mathrm{F}$ is doubled, then new capacity will be
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Verified Answer
The correct answer is:
$5 \mu \mathrm{F}$
In parallel plate capacitor
$$
C=\varepsilon_{0} \frac{A}{d}
$$
Now
$$
d^{\prime}=2 d
$$
Hence,
$$
C^{\prime}=\frac{\varepsilon_{0} A}{d^{\prime}}=\frac{\varepsilon_{0} A}{2 d}
$$
From Eqs. (i) and (ii)
$$
\begin{array}{l}
\frac{C^{\prime}}{C}=\frac{1}{2} \\
C^{\prime}=\frac{C}{2}
\end{array}
$$
Putting $C=10 \mu \mathrm{F}$
$$
C^{\prime}=\frac{10}{2}=5 \mu \mathrm{F}
$$
$$
C=\varepsilon_{0} \frac{A}{d}
$$
Now
$$
d^{\prime}=2 d
$$
Hence,
$$
C^{\prime}=\frac{\varepsilon_{0} A}{d^{\prime}}=\frac{\varepsilon_{0} A}{2 d}
$$
From Eqs. (i) and (ii)
$$
\begin{array}{l}
\frac{C^{\prime}}{C}=\frac{1}{2} \\
C^{\prime}=\frac{C}{2}
\end{array}
$$
Putting $C=10 \mu \mathrm{F}$
$$
C^{\prime}=\frac{10}{2}=5 \mu \mathrm{F}
$$
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