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(C) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$,where $A$ is the area of the plates and $d$

Vedclass · Accepted Answer

$20$

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(C) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.From this relation,we can see that $C \propto \frac{1}{d}$.Therefore,the ratio of the capacitances is $\frac{C_1}{C_2} = \frac{d_2}{d_1}$.Given: $C_1 = 10\,\mu F$,$d_1 = 8\,cm$,and $d_2 = 4\,cm$.Substituting these values into the ratio formula: $\frac{10}{C_2} = \frac{4}{8}$.This simplifies to $\frac{10}{C_2} = \frac{1}{2}$.Solving for $C_2$,we get $C_2 = 10 	imes 2 = 20\,\mu F$.

The capacity of a parallel plate capacitor is $10\,\mu F$,when the distance between its plates is $8\,cm$. If the distance between the plates is reduced to $4\,cm$,then the capacity of this parallel plate capacitor will be.........$\mu F$.

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