Question

An infinite nonconducting sheet has a surface charge density σ = 0.10 µC/m2 on one side. How far apart are equipotential surfaces whose potentials differ by 86 V?

Answer 1

d = 0.152 meters

Step-by-step explanation:

It is given that,

Surface charge density of the sheet,
`\sigma=0.10\ \mu C/m^2=0.1* 10^(-6)\ C/m^2`

Potential, V = 86 V

We know that the electric field due to the non conducting sheet is given by :

`E=(\sigma)/(2\epsilon_o)`

Also, the relation between the electric field and electric potential is given by :

`E=(V)/(d)`

d is the distance between equipotential surfaces

`(V)/(d)=(\sigma)/(2\epsilon_o)`

`d=(2V\epsilon_o)/(\sigma)`

`d=(2* 86* 8.85* 10^(-11))/(0.1* 10^(-6))`

d = 0.152 meters

So, the equipotential surfaces are separated by 0.152 meters. Hence, this is the required solution.