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I have the following question which I have for the most part sussed but I have no idea how to show the last part.

Electric charge is distributed with variable density $\sigma$ on an infinite plane. $P$ is a point at a distance $a$ from the plane, and $dS$ is a surface element whose distance from $P$ is $R$.

$(a)$ Prove that the electric field at P has a component normal to the plane (this is completed ignore)

$(b)$ Show that for constant $σ$, your answer to $(a)$ gives $σ/2ε_0$ (this is completed ignore)

Therefore deduce that in this case $(b)$ one half of the field arises from the points in the plane whose distance from P is less than $2a$. (This the part I don't know where to start. Any help or hints would be greatly appreciated.)

dahaka5
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1 Answers1

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The electric field at $\vec r=\hat z a$, due to a disk of uniform charge density $\sigma$ with radius $\sqrt 3 a$ and located at $\rho <\sqrt 3 a$, $z=0$ is given by

$$\begin{align} \vec E(0,0,a)&=\frac{\sigma}{4\pi \epsilon_0}\int_0^{2\pi}\int_0^{a\sqrt 3} \frac{\hat z a-\hat\rho'\rho'}{\left(a^2+\rho'^2\right)^{3/2}}\,\rho'\,d\rho'\,d\phi'\\\\ &=\hat z \frac{a\sigma }{2\epsilon_0}\int_0^{a\sqrt 3} \frac{\rho'}{(a^2+\rho'^2)^{3/2}}\,d\rho'\\\\ &=\hat z \frac{a\sigma }{2\epsilon_0} \left(\frac{1}{2a}\right)\\\\ &=\hat z \frac{\sigma }{4\epsilon_0} \end{align}$$

which is $1/2$ the electric field at $(0,0,a)$ due to a uniform charge density of $\sigma$, distributed on the plane $z=0$.

And we are done!

Mark Viola
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