Question

An infinite plane of charge has surface charge density 4.7 mu or micro cc/m2. How far apart are the equipotential surfaces whose potentials differ by 100 v?

Answer 1

charge density of infinite sheet is given as

$\sigma = 4.7 \mu C/m^2$

now the electric field due to this charged sheet will be constant and it is given as

$E = \frac{\sigma}{2\epsilon_0}$

$E = \frac{4.7 * 10^{-6}}{2 * 8.85 * 10^{-12}}$

$E = 2.66 * 10^5 N/c$

now the relation between electric field intensity and potential difference is given as

$\Delta V = E.d$

$\Delta V = 100 Volts$

$E = 2.66 * 10^5 N/C$

now we have

$d = \farc{\Delta V}{E}$

$d = \frac{100}{2.66* 10^5}$

$d = 3.76 * 10^{-4}m$

so it is at distance 0.376 mm from the plate