A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles, then (a) the capacitance of the capacitor increases (b) the charge on the capacitor decreases (c) the voltage across the capacitor increases (d) the energy stored in the capacitor decreases

Answer: (c). Battery removed, so Q stays constant. d increases ⇒ C = (ε₀ A)/(d) decreases. V = Q/C increases; U = Q²/2C also increases. So the answer is (c) the voltage across the capacitor increases.

A parallel plate capacitor is charged and then disconnected from the… | MHT CET 2022 (11 Aug, Shift 2)

Q21MHT CET 2022 · 11 AUG · SHIFT 2Easy

Energy Stored & Force between Plates

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles, then

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Answer: (c)

Battery removed, so $Q$ stays constant.
$d$ increases $\Rightarrow C = \frac{\varepsilon_0 A}{d}$ decreases.
$V = \frac{Q}{C}$ increases; $U = \frac{Q^2}{2C}$ also increases.
So the answer is (c) the voltage across the capacitor increases.

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Energy Stored & Force between Plates question from MHT CET 2022 (11 Aug, Shift 2)

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