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Section 11.8 The electric field of a uniform disk

Recall that the electric field on a surface is given by

\begin{gather*} \EE(\rr) = \int \frac{1}{4\pi\epsilon_0} \frac{\sigma(\rrp)(\rr-\rrp)\,dA}{|\rr-\rrp|^3} \end{gather*}

Let's find the electric field due to a charged disk, on the axis of symmetry.

In cylindrical coordinates, we have

\begin{gather*} \rr - \rrp = z\,\zhat - r'\,\rhat\Prime \end{gather*}

(where we write \(\rhat\Prime\) to emphasize that this basis is associated with \(\rrp\)). The integral becomes

\begin{gather*} \EE(z) = \Int_0^{2\pi}\Int_0^R \frac{\sigma}{4\pi\epsilon_0} \frac{(z\,\zhat-r'\,\rhat\Prime)\,r'\,dr'\,d\phi'} {(z^2 + r'^2)^{3/2}} \end{gather*}

It is important to note that \(\rhat\Prime\) can not be pulled out of the integral, since it is not constant. Quite the opposite, by symmetry, this integral must vanish! Explicitly, writing

\begin{gather*} \rhat\Prime = r'\cos\phi'\,\ii + r'\sin\phi'\,\jj \end{gather*}

and then integrating will indeed yield zero. The remaining term is

\begin{align*} \EE(z) \amp= \Int_0^{2\pi}\Int_0^R \frac{\sigma}{4\pi\epsilon_0} \frac{z\,r'\,dr'\,d\phi'} {(z^2 + r'^2)^{3/2}} \> \zhat\\ \amp= -\frac{\sigma\,\zhat}{4\pi\epsilon_0} \frac{2\pi z}{\sqrt{z^2+r'^2}} \Bigg|_0^R = \frac{2\pi\sigma\,\zhat}{4\pi\epsilon_0} \left( \frac{z}{\sqrt{z^2}} - \frac{z}{\sqrt{z^2+R^2}} \right) \end{align*}

Recall that the electric field of a uniform disk is given along the axis by

\begin{gather*} \EE(z) = \frac{2\pi\sigma}{4\pi\epsilon_0} \left( \frac{z}{\sqrt{z^2}} - \frac{z}{\sqrt{z^2+R^2}} \right)\,\zhat \end{gather*}

where of course \(\frac{z}{\sqrt{z^2}}=\pm1\) depending on the sign of \(z\text{.}\) (The notation \({ sgn}(z)\) is often used to represent the sign of \(z\text{,}\) in order to simplify expressions like \(\frac{z}{\sqrt{z^2}}\text{.}\)) In the limit as \(R\to\infty\text{,}\) one gets the electric field of a uniformly charged plane, which is just

\begin{gather*} \EE(z) = \hbox{sgn}(z) \> \frac{\sigma}{2\epsilon_0}\,\zhat \end{gather*}

which is valid everywhere, as any point can be thought of as being on the axis.