Question

The parallel plates of an air capacitor are separated by 2.25 mm. Each plate carries a charge of 6.50 nC. The magnitude of the electric field of the plates is 4.75x10^5 V/m. Find the (a) potential difference between the plates, (b) capacitance, and (c) area of plate.

Answer 1

Solution;

Given;

$d=2.25mm=2.25×10^{-3}m$

$q=6.50nC$

$E=4.75×10^5V/m$

(a) Potential difference;

$V=E×d$

$V=4.75×10^5×2.25×10^{-3}$

$V=1068.75V$

(b) Capacitance;

$C=\frac qV$

$C=\frac{6.50×10^{-9}}{1068.75}=6.082×10^{-12}F$

(c) Area of the plate;

$A=\frac{Cd}{\epsilon_0}$

Where;

$\epsilon_0=8.85×10^{-12}F/m$

$A=\frac{6.082×10^{-12}×2.25×10^{-3}}{8.85×10^{-12}}$

$A=1.546×10^{-3}m^2$