Plate area A = 25cm2 = 2.5 × 10-3m
Separation d = 2mm = 2 × 10-3m
Potential v = 12v
- We know
$\text{C}=\frac{\in_0\text{A}}{\text{d}}$
$=\frac{8.85\times10^{-12}\times2.5\times10^{-3}}{2\times10^{-3}}$
$=11.06\times10^{-12}\text{F}$
$\text{C}=\frac{\text{q}}{\text{v}}$
$\Rightarrow11.06\times10^{-12}=\frac{\text{q}}{12}$
$\Rightarrow\text{q}_1=1.32\times10^{-10}\text{C}.$
- Then d = decreased to 1mm
$\therefore\ \text{d}=1\text{mm}=1\times10^{-3}\text{m}$
$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{\text{q}}{\text{v}}$
$=\frac{8.85\times10^{-12}\times2.5\times10^{-3}}{1\times10^{-3}}=\frac{2}{12}$
$\Rightarrow\text{q}_2=8.85\times2.5\times12\times10^{-12}$
$\Rightarrow\text{q}_2=2.65\times10^{-10}\text{C}.$
$\therefore$ The extra charge given to plate = (2.65 - 1.32) × 10-10 = 1.33 × 10-10C.
