Answer:
The surface charge density is [tex]7.8\times10^{-8} C/m^2[/tex].
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
[tex]V=\frac{6\times 10^{-3}}{8\times 10^{-6}}=750 V[/tex]
Let the charge on he capacitor is q.
[tex]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[/tex]
