Rankers Physics

Capacitors: Practice Problem & Solution

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: (2006)
Decreases
Does not change
Becomes zero
Increases

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

When the battery is disconnected, the charge $Q$ on the plates remains constant. Capacitance $C = \frac{A\varepsilon_0}{d}$. As distance $d$ increases, $C$ decreases. Since $V = \frac{Q}{C}$, a decrease in capacitance results in an increase in potential difference.

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