Answer:
d = 0.152 meters
Explanation:
It is given that,
Surface charge density of the sheet, [tex]\sigma=0.10\ \mu C/m^2=0.1\times 10^{-6}\ C/m^2[/tex]
Potential, V = 86 V
We know that the electric field due to the non conducting sheet is given by :
[tex]E=\dfrac{\sigma}{2\epsilon_o}[/tex]
Also, the relation between the electric field and electric potential is given by :
[tex]E=\dfrac{V}{d}[/tex]
d is the distance between equipotential surfaces
[tex]\dfrac{V}{d}=\dfrac{\sigma}{2\epsilon_o}[/tex]
[tex]d=\dfrac{2V\epsilon_o}{\sigma}[/tex]
[tex]d=\dfrac{2\times 86\times 8.85\times 10^{-11}}{0.1\times 10^{-6}}[/tex]
d = 0.152 meters
So, the equipotential surfaces are separated by 0.152 meters. Hence, this is the required solution.
