Question 24.99: (a) Using Eq. 24-32, show that the electric potential at a p......
(a) Using Eq. 24-32, show that the electric potential at a point on the central axis of a thin ring (of charge q and radius R) and at distance z from the ring is
V=\int d V=\frac{1}{4 \pi \varepsilon_0} \int \frac{d q}{r} . (24-32)
V=\frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{z^2+R^2}} .
(b) From this result, derive an expression for the electric field magnitude E at points on the ring’s axis; compare your result with the calculation of E in Module 22-4.
(a) The charge on every part of the ring is the same distance from any point P on the axis. This distance is r=\sqrt{z^2+R^2} , where R is the radius of the ring and z is the distance from the center of the ring to P. The electric potential at P is
\begin{aligned} V & =\frac{1}{4 \pi \varepsilon_0} \int \frac{d q}{r}=\frac{1}{4 \pi \varepsilon_0} \int \frac{d q}{\sqrt{z^2+R^2}}=\frac{1}{4 \pi \varepsilon_0} \frac{1}{\sqrt{z^2+R^2}} \int d q \\ & =\frac{1}{4 \pi \varepsilon_0} \frac{q}{\sqrt{z^2+R^2}} . \end{aligned}
(b) The electric field is along the axis and its component is given by
\begin{aligned} E & =-\frac{\partial V}{\partial z}=-\frac{q}{4 \pi \varepsilon_0} \frac{\partial}{\partial z}\left(z^2+R^2\right)^{-1 / 2}=\frac{q}{4 \pi \varepsilon_0}\left(\frac{1}{2}\right)\left(z^2+R^2\right)^{-3 / 2}(2 z) \\ & =\frac{q}{4 \pi \varepsilon_0} \frac{z}{\left(z^2+R^2\right)^{3 / 2}} . \end{aligned}
This agrees with Eq. 23-16.
E=0 (spherical shell, field at r < R) (23-16)