(D) Initial capacitance is $C = \frac{\epsilon_0 A}{d}$. Initial charge is $Q_i = CV$.When the distance is reduced to half $(d' = d/2)$,the new capaci

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(D) Initial capacitance is $C = \frac{\epsilon_0 A}{d}$. Initial charge is $Q_i = CV$.When the distance is reduced to half $(d' = d/2)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{d/2} = 2C$.Since the battery remains connected,the potential across the capacitor remains $V$.The new charge on the capacitor is $Q_f = C'V = 2CV$.The additional charge supplied by the battery is $\Delta Q = Q_f - Q_i = 2CV - CV = CV$.The energy supplied by the battery is $W = \Delta Q \times V = (CV) \times V = CV^2$.

Class 12 Physics Electric Potential and Capacitance Mix Examples - Electric Potential and Capacitance

Easy

$A$ capacitor of capacity $C$ is connected to a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once,while the battery remains connected. Then,the energy supplied by the battery to charge the capacitor to potential $V$ again is:

A
$CV^2/4$
B
$CV^2/2$
C
$3CV^2/4$
D
$CV^2$

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Electric Potential and Capacitance

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Mix Examples - Electric Potential and Capacitance

In the adjoining figure,four capacitors are shown with their respective capacities and the $P.D.$ applied. The charge and the $P.D.$ across the $4\,\mu F$ capacitor will be

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The radii of two metallic spheres $P$ and $Q$ are $r_1$ and $r_2$ respectively. They are given the same charge. If $r_1 > r_2$,then on connecting them with a thin wire,the charge will flow

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Minimum number of $2\,\mu F$ and $250\, V$ capacitors required to make a combination of $12\,\mu F$ and $500\, V$ is:

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Two capacitors of capacities $1 \mu F$ and $C \mu F$ are connected in series and the combination is charged to a potential difference of $120 \ V$. If the charge on the combination is $80 \mu C$,the energy stored in the capacitor of capacity $C$ in $\mu J$ is

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In the circuit shown,each capacitor has a capacitance $C$. The emf of the cell is $E$. If the switch $S$ is closed:

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$A$ capacitor of capacity $C$ is connected to a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once,while the battery remains connected. Then,the energy supplied by the battery to charge the capacitor to potential $V$ again is:

Similar Questions

In the adjoining figure,four capacitors are shown with their respective capacities and the $P.D.$ applied. The charge and the $P.D.$ across the $4\,\mu F$ capacitor will be

The radii of two metallic spheres $P$ and $Q$ are $r_1$ and $r_2$ respectively. They are given the same charge. If $r_1 > r_2$,then on connecting them with a thin wire,the charge will flow

Minimum number of $2\,\mu F$ and $250\, V$ capacitors required to make a combination of $12\,\mu F$ and $500\, V$ is:

Two capacitors of capacities $1 \mu F$ and $C \mu F$ are connected in series and the combination is charged to a potential difference of $120 \ V$. If the charge on the combination is $80 \mu C$,the energy stored in the capacitor of capacity $C$ in $\mu J$ is

In the circuit shown,each capacitor has a capacitance $C$. The emf of the cell is $E$. If the switch $S$ is closed:

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