Question

An inverted hemispherical shell consists of two parts: the upper half of a sphere of radius R with its center at the origin of coordinates and a circle in the xy-plane with the same radius R and its center also at the origin. Both parts carry the same uniform surface charge density σ. There is no volume charge density. Find the potential difference between the hemispherical pole (on the z-axis at z=R) and the
origin by direct integration of Coulomb's law for the potential at the two points. Note that result is not zero!

Answer 1

Okay let's see what we can do. Getting the potential by direct integration means
\[ \Phi = \int \vec{E} \cdot d\vec{r} \]
So we would need to find the value for the electric field, right? Any ideas?

Answer 2

\[E=\int\limits_{?}^{?}(k*\sigma dA r)/(\left| r^3 \right|)\]

Answer 3

Uh, no. A good idea would be to calculate the field on the outside of the shell. Then, you could integrate down the z axis to the top to find the potential there, and also up the z axis from the bottom to find the potential there. Obviously then, the potential difference will be the difference of those two.

So we will have to integrate in two parts. First we'll find the field from the disk:

Answer 4

The electric field at a point z on the plate is obviously pointing straight along the z-axis, by symmetry. Therefore, we'll consider only the z-component, which we can get by multiplying the magnitude by z/r (why?)


\[E(z) =k \int \frac{ \sigma }{r^2+z^2} \cdot \frac{z}{r} dA\]
\[ = k\cdot \sigma \cdot z \int_0^{2\pi}d\theta \int_0^R \frac{ dr}{r^2+z^2}\]

Can you do that integral?

Answer 5

yes

Answer 6

Okay, so that's the field from the plate. Now we find the field from the hemispherical shell, which will be a bit more complicated.