Section 11.8 The electric field of a uniform disk
Recall that the electric field on a surface is given by
Let's find the electric field due to a charged disk, on the axis of symmetry.
In cylindrical coordinates, we have
(where we write \(\rhat\Prime\) to emphasize that this basis is associated with \(\rrp\)). The integral becomes
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
and then integrating will indeed yield zero. The remaining term is
Recall that the electric field of a uniform disk is given along the axis by
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
which is valid everywhere, as any point can be thought of as being on the axis.