Question

Equipotential surfaces are to be drawn 100 V apart near a very large uniformly charged metal plate carrying a surface charge density σ = 0.75 μC/m2. How far apart (in space) are the equipotential surfaces?

Answer 1

Electric field due to uniformly charged metal plate is given by,

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

Here,

$\sigma$ = Charge density

$\epsilon_0 =$  Vacuum Permittivity

Our values are,

$\sigma = 0.75 muC/m^2 = 0.75*10^-6 C/m^2$

$\epsilon_0 = 8.85*10^-12 F\cdot m^{-1}$

Replacing we have,

$E = \frac{(0.75*10^-6)}{(2*8.85*10^-12)}$

$F = 42372.88N/C$

Now we have the relation where energy is equal to the change of the potential in a certain distance, then

$E = \frac{V}{d}$

Rearranging for the distance

$d = \frac{V}{E}$

$d = \frac{100}{42372.88}$

$d = 0.00236m$

$d = 2.36mm$

Therefore the distance is 2.36mm