The capacitance of a parallel plate capacitor is $12\,\mu \,F$. If the distance between the plates is doubled and area is halved, then new capacitance will be.........$\mu \,F$
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(d) $C = \frac{{{\varepsilon _0}A}}{d}$ As $A \to \frac{1}{2}$ times and $d \to 2$ times
So $C \to \frac{1}{4}$times i.e. $C' = \frac{1}{4}C = \frac{{12}}{4} = 3\,\mu F$
So $C \to \frac{1}{4}$times i.e. $C' = \frac{1}{4}C = \frac{{12}}{4} = 3\,\mu F$
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