# Find the potential at the center of the sphere.

The inside of a grounded spherical metal shell of radius R is filled with charge of uniform density ρ. Find the potential at the center.

My attempt:

$$V=\frac{1}{4\pi \epsilon_0} \int_{V} \frac{\rho}{d}d \tau \quad ; \quad d=|\overrightarrow r - \overrightarrow {r'}|$$

$$V=\frac{1}{4\pi \epsilon_0} \int_R^0\int_0^\pi\int_0^{2\pi} \frac{\rho}{|0 - \overrightarrow {r'}|}r'^2\sin \theta dr' d\theta d\phi$$

In the line above, I claimed that my field point is at the center and the source point is defined by $r'$, which means I can say that $\overrightarrow r =0$. We are integrating with respect to $r'$.

$$V=\frac{-\rho}{\epsilon_0} \int_R^0 \frac{r'^2}{r'}dr'$$ $$V=\frac{-\rho}{\epsilon_0} \left [ \frac{r'^2}{2}|_R^0 \right ]=\frac{\rho R^2}{2\epsilon_0}$$

...but my professor claims that the answer is

$$V=\frac{\rho R^2}{6\epsilon_0}$$

Where did I make a mistake?

Update:

I asked a tutor and she said there were 2 problems with my derivation. First, I can't integrate from $R$ to $0$. She said this is a regular volume integral and therefore I should do it the 'normal way'. This was pretty poor reasoning, could someone give me a better explanation as to why I can't do this?

Second, she claimed that the '$r$' in $\frac{1}{-\overrightarrow {r}}$ and the one in $d \tau$ are actually different. Why is this?