1 \begin{problem*}{20.24}
2 Consider a ring of radius $R$ with the total charge $Q$ spread
3 uniformly over its perimeter. What is the potential difference
4 between the point at the center of the ring and a point on its axis a
5 distance $d=2R$ from the center?
6 \end{problem*} % problem 20.24
9 From the first week's recitation (P19.19), we have the electric field
10 along the axis due to the ring as
12 \vect{E} = \frac{k_e x Q}{(x^2 + R^2)^{3/2}}\ihat
14 So the potential drop from $0$ to $d$ is given by
16 \Delta V = -\int_0^d E_x dx
17 = - k_e Q \int_0^d \frac{x \cdot dx}{(x^2 + R^2)^{3/2}}
19 Substituting $u = x^2 + R^2$ so $du = 2x dx$ we have
21 \Delta V = - k_e Q \int \frac{1/2 \cdot du}{u^{3/2}}
22 = - \frac{1}{2} k_e Q \frac{-2}{\sqrt{u}}
23 = \frac{k_e Q}{\sqrt{u}}
25 And plugging back in in terms of $x$
27 \Delta V = \left.\frac{k_e Q}{\sqrt{x^2 + R^2}}\right|_0^d
28 = \frac{k_e Q}{\sqrt{d^2 + R^2}} - \frac{k_e Q}{R}
29 = k_e Q \left(\frac{1}{\sqrt{(4+1)R^2}} - \frac{1}{R}\right)
30 = \frac{k_e Q}{R} \left(\frac{1}{\sqrt{5}}-1\right)
31 = \ans{-0.533 \frac{k_e Q}{R}}