(a) Find an expression for the electric potential at a point P located on the perpendicular central axis of a uniformly charged ring of radius a and total charge Q.
Question : (a) Find an expression for the electric potential at a point...
Let us orient the ring so that its plane is perpendicular to an x axis and its center is at the origin. We can then take point P to be at a distance x from the center of the ring, as shown in Figure 25.15 . The charge element d q is at a distance \sqrt{x^{2}+a^{2}} from point P. Hence, we can express \operatorname{Vas}
V=k_{e} \int \frac{d q}{r}=k_{e} \int \frac{d q}{\sqrt{x^{2}+a^{2}}}Because each element dq is at the same distance from point P_{z}we can remove \sqrt{x^{2}+a^{2}} from the integral, and V reduces to
V=\frac{k_{\ell}}{\sqrt{x^{2}+a^{2}}} \int d q=\frac{k_{e} Q}{\sqrt{x^{2}+a^{2}}}The only variable in this expression for V is x. This is not surprising because our calculation is valid only for points along the xaxis, where y and z are both zero.
(b) Find an expression for the magnitude of the electric field at point P.
From symmetry, we see that along the x axis \mathbf{E} can have only an x component. Therefore, we can use Equa-tion 25.16
E_{x} =-\frac{d V}{d x}=-k_{e} Q \frac{d}{d x}\left(x^{2}+a^{2}\right)^{-1 / 2} =-k_{e} Q\left(-\frac{1}{2}\right)\left(x^{2}+a^{2}\right)^{-3 / 2}(2 x) =\frac{k_{e} Q x}{\left(x^{2}+a^{2}\right)^{3 / 2}}This result agrees with that obtained by direct integration (see Example 29.8). Note that E_{x}=0 at x=0 (the center of the ring).