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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 :-
A
Decreases
B
Does not change
C
Becomes zero
D
Increases
Detailed Solution
After the battery is disconnected the plates are isolated, so the charge $Q$ on the capacitor remains constant.
Capacitance of a parallel plate capacitor: $C = \dfrac{\varepsilon_0 A}{d}$
When the distance $d$ between the plates is increased, the capacitance $C$ decreases.
Potential difference: $V = \dfrac{Q}{C}$
Since $Q$ is constant and $C$ decreases, $V$ increases.
(The field $E = \dfrac{Q}{\varepsilon_0 A}$ stays the same, and $V = Ed$ increases with $d$.)
Capacitance of a parallel plate capacitor: $C = \dfrac{\varepsilon_0 A}{d}$
When the distance $d$ between the plates is increased, the capacitance $C$ decreases.
Potential difference: $V = \dfrac{Q}{C}$
Since $Q$ is constant and $C$ decreases, $V$ increases.
(The field $E = \dfrac{Q}{\varepsilon_0 A}$ stays the same, and $V = Ed$ increases with $d$.)