# Potential of a uniformly charged hollow sphere [closed]

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 ?

## closed as off-topic by ACuriousMind♦, Kyle Kanos, JamalS, Jim, Constandinos DamalasApr 9 '15 at 18:47

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• physics.stackexchange.com/a/174609/75518 – image Apr 7 '15 at 14:22
• Why do I need a dirac delta ? What is wrong with what I wrote ? – mwa1 Apr 7 '15 at 15:13
• If you just do the integral over $r'$ you'll get the integral over the surface of your sphere. The way I solved it, is just a way to show you, where this surface integration comes from and that one can forumalte a proper charge density for this kind of problem. The dirac-delta ensures, that the charge is only on the surface – image Apr 7 '15 at 15:31
• I still don't understand what went wrong with my developement... Also, what does $d^3r'$ mean ? I don't understand this notation – mwa1 Apr 7 '15 at 19:57
• $\int f \ \ d^3r\ '$ denotes a volume integral over $f$ independant of the choise of coordinates. – image Apr 7 '15 at 20:17

## 1 Answer

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 $\phi$.

• Ok, thanks. The $1/r$ factor was a typo, I got rid of it. – mwa1 Apr 7 '15 at 20:21
• Ok, good! I'll take that part out of my answer. – Mike Bell Apr 7 '15 at 20:22