Skip to main content Contents Index Prev Up Next \(\newcommand{\vf}[1]{\mathbf{\boldsymbol{\vec{#1}}}}
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\)
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.
