Question

A parallel-plate air capacitor has a capacitance of $4\mu F$. What will be its new capacitance if the distance between the plates is reduced to half the initial distance ?

Answer 1

Capacitance of a parallel plate capacitor is given by:

$C=\frac{{\epsilon }_{0}A}{d}$

Now distance is halved i.e $d/2$

New capacitance is,

$C_1=\frac{{\epsilon }_{0}A}{d/2}=\frac{2{\epsilon }_{0}A}{d}$

$C_1=2C=2\times 4\mu F=8\mu F$