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A capacitor of capacitance $C_{0}$ is charged to potential $4 V .$ The flux of electric field through a closed surface enclosing the positive plate is $\frac{C V}{ε_{0}}$ then value of $C_{0}$ in terms of $C$ is :

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$C_{0} = 4 C$

$C_{0} = 2 C$

$C_{0} = \frac{C}{2}$

$C_{0} = \frac{C}{4}$

answer is A.

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Detailed Solution

$Q = 4 C_{0} V ϕ = \frac{4 C_{0} V}{ε_{0}} = \frac{C V}{ε_{0}} 4 C_{0} = C C_{0} = \frac{C}{4}$
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