Question 24.93: A uniform charge of +16.0 μC is on a thin circular ring lyin......
A uniform charge of +16.0 μC is on a thin circular ring lying in an xy plane and centered on the origin. The ring’s radius is 3.00 cm. If point A is at the origin and point B is on the z axis at z = 4.00 cm, what is V_B-V_A?
THINK To calculate the potential at point B due to the charged ring, we note that all points on the ring are at the same distance from B.
EXPRESS Let point B be at (0, 0, z). The electric potential at B is given by
V=\frac{q}{4 \pi \varepsilon_0 \sqrt{z^2+R^2}}
where q is the charge on the ring. The potential at infinity is taken to be zero.
ANALYZE With q=16 \times 10^{-6} \,C , z=0.040 \,m and R = 0.0300 m, we find the potential difference between points A (located at the origin) and B to be
\begin{aligned} V_B-V_A & =\frac{q}{4 \pi \varepsilon_0}\left(\frac{1}{\sqrt{z^2+R^2}}-\frac{1}{R}\right) \\ & =\left(8.99 \times 10^9 \,N \cdot m ^2 / C ^2\right)\left(16.0 \times 10^{-6}\, C \right)\left(\frac{1}{\sqrt{(0.030\, m )^2+(0.040\, m )^2}}-\frac{1}{0.030 \,m }\right) \\ & =-1.92 \times 10^6 \,V . \end{aligned}
LEARN In the limit z ≫ R, the potential approaches its “point-charge” limit:
V \approx \frac{q}{4 \pi \varepsilon_0 z} .