Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

decreases.

does not change.

becomes zero.

increases.

Step-by-Step Solution

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.
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 $C$ .
Potential Difference: The potential difference is related to charge and capacitance by $V = \frac{Q}{C}$ [2].
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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