A parallel-plate capacitor has plate area $25.0c{m}^2$ and a separation of 2.00 mm between the plates. The capacitor is connected to a battery of 12.0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1.00 mm. Find the extra charge given by the battery to the positive plate.

plate are $A=25c{m}^2$

$=25\times {10}^{-4}{m}^2,$

separation $d=2mm=2\times {10}^{-3}m$

potential $v=12V$

(a) Capacitance $C=\frac{{\in }_{0}A}{d}$

$=\frac{8.88\times {10}^{-12}\times 25\times {10}^{-4}}{2\times {10}^{-3}}$

$=11.06\times {10}^{-12}F$

charge $q_1=CV$

$=11.06\times {10}^{-12}\times 12$

$=1.32\times {10}^{-10}C$

(b) $d=1mm=1\times {10}^{-8}m$

So, $C=\frac{{\in }_{0}A}{d}$

$=\frac{8.88\times 2\times {10}^{-3}}{1\times {10}^{-12}}$

$=22.06\times {10}^{-12}F$

$q_2=CV$

$={22.06}^{\prime }\times {10}^{-12}\times 12$

$=2.65\times {10}^{-10}C$

Extra charge

$=(2.65\times {10}^{-10}-1.32\times {10}^{-10})C$

$=1.33\times {10}^{-10}C$