A parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the char

Question

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:

Accepted Answer

1. **Conservation of Charge:** When the charging battery is disconnected, the charge $Q$ stored on the capacitor plates remains constant because it is isolated from the circuit.
2. **Capacitance Formula:** The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$, where $d$ is the separation distance [1]. Increasing the distance $d$ decreases the capacitance $C$.
3. **Potential Difference:** The potential difference is related to charge and capacitance by $V = \frac{Q}{C}$ [2].
4. **Conclusion:** Since $Q$ is constant and $C$ decreases (due to the increase in $d$), the denominator in the expression for $V$ becomes smaller, causing the potential difference $V$ to **increase**.

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