A capacitor of capacitance $C$ has charge $Q$ and resultant energy stored in it is $W$ . If the charge on its plates is increased to $2 Q$ , the stored energy would be

\frac{W}{2}

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2 W

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\frac{W}{4}

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4 W

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Solution

The correct option is D $4 W$
Enery stored when capactitor has charge $Q$ is $E = W = \frac{Q^{2}}{2 C}$

Enery stored when capactitor has charge $2 Q$ is
$W^{1} = \frac{(2 Q)^{2}}{2 C} = 4 (\frac{Q^{2}}{2 C})$
$W^{1} = 4 W$

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A capacitor of capacitance C has charge Q and resultant energy stored in it is W. If the charge on its plates is increased to 2Q, the stored energy would be

The correct option is D 4WEnery stored when capactitor has charge Q is E=W=Q22C Enery stored when capactitor has charge 2Q is W1=(2Q)22C=4(Q22C) W1=4W

A capacitor of capacitance $C$ has charge $Q$ and resultant energy stored in it is $W$ . If the charge on its plates is increased to $2 Q$ , the stored energy would be

The correct option is D $4 W$
Enery stored when capactitor has charge $Q$ is $E = W = \frac{Q^{2}}{2 C}$

Enery stored when capactitor has charge $2 Q$ is
$W^{1} = \frac{(2 Q)^{2}}{2 C} = 4 (\frac{Q^{2}}{2 C})$
$W^{1} = 4 W$