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A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then
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Verified Answer
The correct answer is:
the voltage across the capacitor increases
We know that
The parallel plate capacitor $C=\frac{\varepsilon_{0} \cdot A}{d}$
When $d$ (distance between two parallel plate) increase, then $C$ will be decrease
$\because$
$Q=C V \quad$ where $Q=$ constant
$\therefore$ The voltage across the capacitor increases.
The parallel plate capacitor $C=\frac{\varepsilon_{0} \cdot A}{d}$
When $d$ (distance between two parallel plate) increase, then $C$ will be decrease
$\because$
$Q=C V \quad$ where $Q=$ constant
$\therefore$ The voltage across the capacitor increases.
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