Question

Two large parallel metal plates carry opposite charges. They are separated by 85mm. The work done by the field is 6x10-3J and its field exerts on a particle with charge +8µC. Calculate the surface charge density on each plate.

Answer 1

Answer:

The surface charge density is $7.8\times10^{-8} C/m^2$.

Step-by-step explanation:

separation, d = 85 mm

Work, W = 6 x 10^-3 J

charge , Q = 8µC

The potential difference is given by

W = q V

$V=\frac{6\times 10^{-3}}{8\times 10^{-6}}=750 V$

Let the charge on he capacitor is q.

$q = CV\\\\q = \frac{\varepsilon oA}{d}\times V\\\\\frac{q}{A} = \frac{8.85\times 10^{-12}\times750}{0.085} =7.8\times10^{-8} C/m^2$