Potential of a uniformly charged hollow sphere [closed]

Question

If I use Gauss' theorem I find that, for $r\geq R$

$$V(r) = \frac{\sigma R^2}{\epsilon_0 r} = \frac{Q_{sphere}}{4\pi\epsilon_0r}$$

where $\sigma$ is the surface charge density and $R$ the radius of the sphere.

But if I use the "direct" formula, I find that

$$V(r) = \frac{1}{4\pi\epsilon_0}\iint_{sphere}\frac{\sigma(P)}{||\vec{PM}||}dS$$

with $P\in sphere$

and since the sphere is uniformly charged and $||\vec{PM}|| = |r-R|$

$$V(r) = \frac{\sigma}{4\pi\epsilon_0|r-R|}\iint_{sphere}dS = \frac{Q_{sphere}}{4\pi\epsilon_0|r-R|}$$

What am I doing wrong ?

Answer 1

The "direct" formula is $$V(r)=\frac{1}{4\pi\epsilon_0}\int\frac{dQ}{\lvert \vec{r}-\vec{R} \rvert}=\frac{1}{4\pi\epsilon_0}\iint_{sphere}\frac{\sigma(\vec{R})dS}{\lvert \vec{r}-\vec{R} \rvert}.$$

Now, think carefully about what the $\frac{1}{\lvert \vec{r}-\vec{R} \rvert}$ means---it is the reciprocal of the distance from an arbitrary point on the surface of the sphere to the point where you are calculating the electrostatic potential. So as you integrate over the surface of the sphere, this distance changes, and so you cannot pull it out of the integral as you did above. To actually evaluate the integral, you would need to write $dS$ and $\frac{1}{\lvert \vec{r}-\vec{R} \rvert}$ in terms of, for instance, the polar coordinates $\theta$ and $\phi$.