Question

An infinite sheet of charge has a surfaces charge density of 10^-7C/m^2. The separation between two equipotential surfce whose potentials differ by 5 V is

Answer 1

Given:

An infinite sheet of charge has a surfaces charge density of 10^-7C/m^2.

Potential difference between the 2 equipotential surface is 5 V.

To Find:

The separation between two equipotential surface.

Solution:

We know , Electric field due to infinite charge sheet, E is given by ,

• E = $\frac{\sigma}{2\epsilon_{0} }$

Also Potential difference and Electric field is related by ,

• V = E x d,
• Where d is the separation between two surfaces.

Therefore combining both the equations,

• V/d = $\frac{\sigma}{2\epsilon_{0} }$ =
• d = 2V$\epsilon_{0}$ / $\sigma$ = 2x 5 x 8.85 x $10^{-12}$ / $10^{-7}$ = 8.85 x $10^{-4}$ m
• d = 0.885mm

Therefore, the separation between two equipotential surface is 0.885mm