Question

The capacitance of a capacitor is 10 μF. The distance between the plates is made half and area is doubled. The capacitance of the capacitor will now be:

A. 5 μF
B.10 μF
C.20 μF
D.40 μF

Answer 1

`C=(\epsilon_(0). \alpha)/(d), \alpha=area,~d=distence~b/w~plates\\so~ C \propto \left ( (\alpha)/(d) \right )\\(C_(2))/(C_(1)) = \left ( (\alpha_(2))/(\alpha_(1)) \right ) . \left ( (d_(1))/(d_(2)) \right ) \\ given~ \alpha_(2)=2.\alpha_(1)~and, d_(2)=(d_(1))/(2)\\ \boxed{C_(2)=4.C_(1)}`
so capacitence becomes 4 times i.e 40uF