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 $C = 12\,\mu F$.Wh

Vedclass · Accepted Answer

$3$

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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 distance between the plates is doubled,$d' = 2d$.When the area of the plates is halved,$A' = \frac{A}{2}$.The new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A'}{d'} = \frac{\varepsilon_0 (A/2)}{2d} = \frac{1}{4} \left( \frac{\varepsilon_0 A}{d} ight) = \frac{1}{4} C$.Substituting the value of $C$,we get $C' = \frac{1}{4} 	imes 12\,\mu F = 3\,\mu F$.

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

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