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 = (\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 = ((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 = (V)/(d)`

Rearranging for the distance

`d = (V)/(E)`

`d = (100)/(42372.88)`

`d = 0.00236m`

`d = 2.36mm`

Therefore the distance is 2.36mm