Question 24.102: A charge of 1.50 × 10^-8 C lies on an isolated metal sphere ......
A charge of 1.50 × 10^{-8} C lies on an isolated metal sphere of radius 16.0 cm. With V = 0 at infinity, what is the electric potential at points on the sphere’s surface?
We imagine moving all the charges on the surface of the sphere to the center of the the sphere. Using Gauss’ law, we see that this would not change the electric field outside the sphere.
The magnitude of the electric field E of the uniformly charged sphere as a function of r, the distance from the center of the sphere, is thus given by E(r)=q /\left(4 \pi \varepsilon_0 r^2\right) \,\,\text { for } r>R .
Here R is the radius of the sphere. Thus, the potential V at the surface of the sphere (where r = R) is given by
\begin{aligned} V(R) & =\left.V\right|_{r=\infty}+\int_R^{\infty} E(r) d r=\int_{\infty}^R \frac{q}{4 \pi \varepsilon_0 r^2} d r=\frac{q}{4 \pi \varepsilon_0 R}=\frac{\left(8.99 \times 10^9 \frac{ N \cdot m ^2}{ C ^2}\right)\left(1.50 \times 10^8 C \right)}{0.160 \,m } \\ & =8.43 \times 10^2\, V . \end{aligned}