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Charges on Conductors

A parallel pl...

Question

A parallel plate capacitor $C$ carries a charge $Q$ . The distance between the plates is doubled by application of force, the work done by force is

z e r o

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\frac{Q^{2}}{2 C}

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\frac{Q^{2}}{C}

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\frac{Q^{2}}{4 C}

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Solution

The correct option is B $\frac{Q^{2}}{2 C}$
Energy before separation,

$E = \frac{Q^{2}}{2 C}$

Energy after separation,

$E = \frac{Q^{2}}{C}$

Work done is difference in energy

$W = \frac{Q^{2}}{2 C}$

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Similar questions

(C = 6 n F)

Q_{1} = 12 n C

Q_{2} = 6 n C

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\frac{9 Q^{2}}{8 C}

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\frac{3 Q}{2 C}

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\frac{Q^{2}}{A ϵ_{0}}

(C = 6 n F)

Q_{1} = 12 n C

Q_{2} = 6 n C

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