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

The surface charge density is
`7.8*10^(-8) C/m^2`.

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=(6* 10^(-3))/(8* 10^(-6))=750 V`

Let the charge on he capacitor is q.

`q = CV\\\\q = (\varepsilon oA)/(d)* V\\\\(q)/(A) = (8.85* 10^(-12)*750)/(0.085) =7.8*10^(-8) C/m^2`