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## Problem

Find the electric ﬁeld a distance $x$ along the axis from a disc of radius $R$ and uniform charge density $\sigma$.
Hint: a disk can be thought of as a bunch of concentric rings. Begin by ﬁnding the electric field a distance $x$ along the axis up from a thin ring of charge $dQ$ and radius $a$.

You may use the integral

$\int \frac{xdx}{(\sqrt{x^2 + a^2})^3} = -\frac{1}{\sqrt{x^2 + a^2}} + C$.

## Solution

Consider the ring problem first. Due to the symmetry of this geometry, there is a cancellation effect in the y-direction. Therefore, all contributions to the electric field in the x-direction. The distance of a point on the x-axis from the ring is $r = \sqrt{x^2 + a^2}$.

Hence, the differential element of electric field (along the x-axis) is

$dE_x = dE \cos{(\theta)}$
$dE_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{dQ}{x^2 + a^2} \frac{x}{\sqrt{x^2 + a^2}}$.

Integrating yields,

$E_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{x}{(\sqrt{x^2 + a^2})^3} \int_{0}^{Q} dQ$
$E = \frac{1}{4\pi {\epsilon}_{0}} \frac{xQ}{(\sqrt{x^2 + a^2})^3} \boldsymbol{i}$.

For the disk problem, replace the differential charge with $dQ = \sigma 2\pi r dr$ because the areal charge density is defined as $\sigma = \frac{dQ}{dA}$ and $A = \pi r^2$ for a disk.

Therefore, the electric field is modified to (we are integrating with respect to the variable $r$ now, so $a$ is replaced with $r$)

$dE_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{\sigma 2\pi rdr}{x^2 + r^2} \frac{x}{\sqrt{x^2 + r^2}}$.

Using the given integral yields,

$E_x = \frac{1}{4\pi {\epsilon}_{0}} \int_{0}^{R} \frac{(\sigma 2\pi rdr) x}{(\sqrt{x^2 + r^2})^3}$
$E_x = \frac{\sigma x}{2 {\epsilon}_{0}} \left( -\frac{1}{\sqrt{x^2 + r^2}} + \frac{1}{x} \right)$.

Therefore,

$E = \frac{\sigma}{2 {\epsilon}_{0}} \left(1 -\frac{x}{\sqrt{x^2 + r^2}} \right) \boldsymbol{i}$.
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