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A parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:
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Verified Answer
The correct answer is:
increases
According to the question $\phi=$ Const and decrease in $C$ means increase in $V$.
$$
\begin{aligned}
& C=\frac{t o f}{d} \\
& \text { Capacitance }=\frac{\text { Charge }}{\text { Potential difference }} \\
& \Rightarrow \text { Potential difference } \\
& =\frac{\text { Charge }}{\text { Capacitance }} \text {. } \\
&
\end{aligned}
$$
$$
\begin{aligned}
& C=\frac{t o f}{d} \\
& \text { Capacitance }=\frac{\text { Charge }}{\text { Potential difference }} \\
& \Rightarrow \text { Potential difference } \\
& =\frac{\text { Charge }}{\text { Capacitance }} \text {. } \\
&
\end{aligned}
$$
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