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(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$.From this,we can see that $C \propto \frac{1}{d

Vedclass · Accepted Answer

$45$

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(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$.From this,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: $C_1 = 15\ \mu F$,$d_1 = 6\, cm$,and $d_2 = 2\, cm$.Substituting these values into the ratio formula:$\frac{15}{C_2} = \frac{2}{6}$$\frac{15}{C_2} = \frac{1}{3}$$C_2 = 15 	imes 3 = 45\ \mu F$.Thus,the new capacitance is $45\ \mu F$.

When the distance between the two plates of a parallel plate capacitor is $6\, cm$,its capacitance is $15\ \mu F$. If the distance between the plates is reduced to $2\, cm$,the capacitance of the parallel plate capacitor becomes ... $\mu F$.

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