The capacitance of a parallel plate capacitor | vedclass

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(D) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$.Given initial capacitance $C = 12 \ \mu F$.W

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$3$

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(D) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$.Given initial capacitance $C = 12 \ \mu F$.When the area $A$ is halved $(A' = A/2)$ and the distance $d$ is doubled $(d' = 2d)$,the new capacitance $C'$ becomes:$C' = \frac{\varepsilon_0 A'}{d'} = \frac{\varepsilon_0 (A/2)}{2d} = \frac{1}{4} \left( \frac{\varepsilon_0 A}{d} \right)$.Substituting the initial value of $C$:$C' = \frac{1}{4} C = \frac{12 \ \mu F}{4} = 3 \ \mu F$.

The capacitance of a parallel plate capacitor is $12 \ \mu F$. If the distance between the plates is doubled and the area of the plates is halved,then the new capacitance will be ... $\mu F$.

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