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Question

An infinite nonconducting sheet has a surface charge density $$\sigma=0.10 \ \mu C/m^2$$ on one side. How far apart are equipotential surfaces whose potentials differ by $$50 \ V$$?

Solution
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The electric field produced by an infinite sheet of charge has magnitude $$E= \dfrac{\sigma}{2 \epsilon_0},$$ where $$\sigma$$ is the surface charge density. The field is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in the direction of the field. Then the electric potential is

$$ V = V_s - \int^x_0 E \ dx =V_s-Ex, $$

where $$V_s$$ is the potential at the sheet. The equipotential surfaces are surfaces of constant $$x$$; that is, they are planes that are parallel to the plane of charge. If two surfaces are separated by $$\Delta x$$ then their potentials differ in magnitude by
$$ \Delta V=E \Delta x = \bigg( \dfrac{\sigma}{2 \epsilon_0} \bigg) \Delta x $$
Thus,
$$ \Delta x= \dfrac{2 \epsilon_0 \Delta V}{\sigma} = \dfrac{2(8.85 \times 10^{-12} \ C^2/N.m^2)(50 \ V)}{0.10 \times 10^{-6} \ C/m^2}=8.8 \times 10^{-3} \ m$$

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