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## Homework Statement

Consider the spherical shell of radius R shown in the figure characterized by a hemisphere of surface charge density [itex]\sigma[/itex] and total charge q in the upper half plane and a hemisphere of surface charge density -[itex]\sigma[/itex] and total charge -q in the lower half plane. Find the electric potential at:

(a) [itex]\vec{x}[/itex]=(7R,14R,0)

(b) and at [itex]\vec{x}[/itex] = [itex]\vec{0}[/itex]

## Homework Equations

General form for a volume charge: V( [itex]\vec{r}[/itex]) = k [itex]\int\frac{\rho( \textbf{r'})}{<r>}d\tau'[/itex]

<r> is supposed to be a script r but I can't figure out how to make that worth in latex, sorry if that's confusing and I am open to suggestions!

## The Attempt at a Solution

I believe (b) to be zero by observation. I am having trouble with (a).

-My first assumption is that the positive half is 'above' the x,y plane and the negative half is 'below' the x,y plane. It isn't made clear but it is not arbitrary considering I am given coordinates to analyze on the x,y plane. There is not a coordinate system or set of axis drawn in the diagram. Also, this is a little confusing because the point of interest is given as [itex]\vec{x}[/itex] and the convention that follows is a vector of magnitude x in the direction of x hat. He has stressed such conventions in other problems.

-I looked for similar examples to extrapolate from but the ones I find explicitly place the point of interest on the axis of symmetry (positive z, usually).

-If my first assumption is correct then the point of interest is on the x,y plane but all that really matters is the magnitude of the vector given (distance from origin) because the charge distribution is symmetric about the z axis. In other words, it doesn't matter if I rotate around the z in the x,y plane but does matter how far away I am. The x and y contributions of the positive cancel those of the negative and we are left with contributions in the z direction.

I tried to use this:

V( [itex]\vec{r}[/itex]) = k [itex]\int\frac{\rho( \textbf{r'})}{<r>}d\tau'[/itex]

to brute force integrate the charges separately and utilize superposition to add them together. I set it up like you would for a sphere (several examples available) but only integrated from pi/2 to 0 (starting on the x, y plane and rotating up to the z axis) so I get a half sphere instead. I have a result but I really can't tell if it makes any sense. I will happily post more on this if this route is reasonably suited for this type of problem.

To further complicate matters we did just learn about multipole expansions today but this problem doesn't ask for an approximation of the potential at large r. So I have to assume my prof wants an exact answer unless an exact answer is unreasonably difficult to come by.

Sorry for the wall of text but I am struggling with my intuition in this course and hope to be corrected where it fails. I am open to any suggestions. Thank you for your help!

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