**Required math: calculus**

**Required physics: electrostatics**

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 3.39.

As another example of applying the solution to Laplace’s equation in cylindrical coordinates, we consider the following problem. We are given a cylindrical non-conducting shell or radius carrying a charge density of on its upper half () and on its lower half (. Find the potential everywhere.

We can begin in the same manner as for the other problem involving a cylindrical shell that we solved earlier. The solution is the same up until the point where we introduce the surface charge.

Thus, outside the shell, we have

Inside the shell, we have

Since the potential is continuous over a surface charge, we must have , so we get

Equating coefficients of the sine and cosine, we get

The outward derivative of the potential is discontinuous across a surface charge, and we have

Plugging in the formulas for and , we get

Since the surface charge is an odd function of , we can eliminate the cosine terms, since the cosine is an even function. Therefore, we have . We are free to choose the potential at infinity to be any constant, so we might as well take it to be zero, in which case we have . We are therefore left with, after simplifying the term in brackets:

To find the , we can use the fact that the set of functions is orthogonal over the interval . That is

Therefore we can multiply both sides by and integrate to get

The integrals in brackets on the right come out to

Therefore (changing the index from to for convenience):

We thus get

The final formula for the potential is

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## Comments

greens theorem would work too.

shouldn’t dV/dr = -nR^(2n)/(R^(n-1)) not R^(n+1)?

The derivative is

so I think the formula is correct as it stands. Remember that