Answered

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:

A5 μF

B.10 μF

C.20 μF

D.40 μF

Verified Answer :

naveen1997

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

Related Questions