1.6F: Field of a Uniformly Charged Infinite Plane Sheet

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Jun 20, 2021
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- 1.6E: Field on the Axis of a Uniformly Charged Disc
- 1.7: Electric Field D

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All we have to do is to put $α = π/2$ in equation 1.6.10 to obtain

$E=\frac{\sigma}{2\epsilon_0}.\tag{1.6.12}$

This is independent of the distance of P from the infinite charged sheet. The electric field lines are uniform parallel lines extending to infinity.

Summary

$\begin{align}&\text{Point charge Q :}\quad \quad \quad &&E=\frac{Q}{4\pi\epsilon_0 r^2}. \\ &\text{Hollow Spherical Shell: } &&E=\text{ zero inside the shell,} \\ & &&E=\frac{Q}{4\pi\epsilon_0 r^2}\text{ outside the shell} \\ &\text{Infinite charged rod :} &&E=\frac{\lambda}{2\pi\epsilon_0 r}. \\ &\text{Infinite plane sheet :} &&E=\frac{\sigma}{2\epsilon_0}. \end{align}$

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