(A) For a parallel plate capacitor,the capacitance is given by $C = \frac{\epsilon_0 A}{d}$.Assuming the capacitor is disconnected from the battery,th

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The potential difference of the capacitor will become $180 \ V$

(A) For a parallel plate capacitor,the capacitance is given by $C = \frac{\epsilon_0 A}{d}$.Assuming the capacitor is disconnected from the battery,the charge $Q$ remains constant.Since $Q = CV$,we have $V = \frac{Q}{C} = \frac{Qd}{\epsilon_0 A}$.This implies $V \propto d$.Therefore,$\frac{V_1}{V_2} = \frac{d_1}{d_2}$.Given $V_1 = 60 \ V$,$d_1 = 4 \ mm$,and $d_2 = 12 \ mm$.$V_2 = \frac{V_1 \times d_2}{d_1} = \frac{60 \times 12}{4} = 180 \ V$.

Class 12 Physics Electric Potential and Capacitance Parallel Plate Capacitor

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The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

A
The potential difference of the capacitor will become $180 \ V$
B
The $P.D.$ will become $20 \ V$
C
The $P.D.$ will remain unchanged
D
The charge on the capacitor will reduce to one third

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Electric Potential and Capacitance

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Parallel Plate Capacitor

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The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

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