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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 3.28

Here's a representation of the potential (consider a cross section of the cylinder)

Let's recall the general solution to Laplace's equation in Cylindrical coordinates (with $z$-symmetry):

$$V(s,\varphi )=a_0+b_0\ln s+\sum _{k=1}^{\mathrm{\infty }}(s^k[a_k\cos \left(k\varphi \right)+b_k\sin \left(k\varphi \right)]+s^{-k}[c_k\cos \left(k\varphi \right)+d_k\sin \left(k\varphi \right)])$$

Let's see how we can start to clean up this expression:

Inside, we get rid of the $\ln s$ term and the terms with $s^{-k}$, as these will blow up as $s\to 0$. At this point the general solution becomes

$$V_{\text{inside}}(s,\varphi )=a_0+\sum _{k=1}^{\mathrm{\infty }}(s^k[a_k\cos \left(k\varphi \right)+b_k\sin \left(k\varphi \right)])$$

Outside, we get rid of the $\ln s$ term and the terms with $s^k$, as these will blow up as $s\to \mathrm{\infty }$. At this point the general solution becomes

$$V_{\text{outside}}(s,\varphi )=a_0+\sum _{k=1}^{\mathrm{\infty }}(s^{-k}[c_k\cos \left(k\varphi \right)+d_k\sin \left(k\varphi \right)])$$

Let's now recall section 2.3.5: Boundary Conditions, where we learned that the potential $V$ and surface charges are related by:

$$\begin{array}{}\text{(2.36, page 88)}& \frac{\partial V_{\text{above}}}{\partial n}-\frac{\partial V_{\text{below}}}{\partial n}=-\frac{1}{{ϵ}_0}\sigma \end{array}$$

It is clear that (where above is now outside and below is inside)

$$\begin{array}{rl}\frac{\partial V_{\text{outside}}}{\partial s}& =\sum _{k=1}^{\mathrm{\infty }}(-k{s}^{-k-1}[c_k\cos \left(k\varphi \right)+d_k\sin \left(k\varphi \right)])\\ \frac{\partial V_{\text{inside}}}{\partial s}& =\sum _{k=1}^{\mathrm{\infty }}(k{s}^{k-1}[a_k\cos \left(k\varphi \right)+b_k\sin \left(k\varphi \right)])\end{array}$$

We know what $\sigma$ is (given by the problem), so:

$$\begin{array}{rl}\sigma =a\sin \left(5\varphi \right)& =-{ϵ}_0(\sum _{k=1}^{\mathrm{\infty }}k{s}^{k-1}[a_k\cos \left(k\varphi \right)+b_k\sin \left(k\varphi \right)]-k{s}^{-k-1}[c_k\cos \left(k\varphi \right)+d_k\sin \left(k\varphi \right)]){|}_{s=R}\\ & =-{ϵ}_0(\sum _{k=1}^{\mathrm{\infty }}k{R}^{k-1}[a_k\cos \left(k\varphi \right)+b_k\sin \left(k\varphi \right)]-k{R}^{-k-1}[c_k\cos \left(k\varphi \right)+d_k\sin \left(k\varphi \right)])\\ & =-{ϵ}_0(\sum _{k=1}^{\mathrm{\infty }}k{R}^{k-1}[a_k\cos \left(k\varphi \right)+b_k\sin \left(k\varphi \right)]-\frac{k}{R^{k+1}}[c_k\cos \left(k\varphi \right)+d_k\sin \left(k\varphi \right)])\end{array}$$

We have a sine on the left hand side, and a sum of different sines and cosines on the right hand side. It is clear that a lot of things will have to go away. We start by getting rid of all the cosines, so:

$$\begin{array}{rl}{a}_k& =0;\text{for all }k\\ c_k& =0;\text{for all }k\end{array}$$

If we have a $\sin \left(5\varphi \right)$ cosine on the left hand side, then we only want that one to be on the right hand side. This means that

$$\begin{array}{rl}{b}_k& =0;\text{whenever }k\ne 5\\ d_k& =0;\text{whenever }k\ne 5\end{array}$$

We are left with

$$a\sin \left(5\varphi \right)={ϵ}_0(5R^{5-1}{b}_5-\frac{5}{R^{5+1}}{d}_5)\sin \left(5\varphi \right)$$

So

$$a=5{ϵ}_0(R^4{b}_5+\frac{1}{R^6}{d}_5)$$

So far we have

$$\begin{array}{rl}{V}_{\text{inside}}(s,\varphi )& =a_0+s^5{b}_5\sin \left(5\varphi \right)\\ V_{\text{outside}}& =a_0+s^{-5}{d}_5\sin \left(5\varphi \right)\end{array}$$

But we are not done yet. We need another equation to relate $b_5$ and $d_5$ to get explicit values for each. An additional relation between them comes from the fact that $V$ must be continuous where both parts meet ($s=R$), that is:

$$V_{\text{inside}}(R,\varphi )=V_{\text{outside}}(R,\varphi )$$

It is clear at this point that we can ignore $a_0$ completely, and that

$$s^5{b}_5\sin \left(5\varphi \right)=s^{-5}{d}_5\sin \left(5\varphi \right)$$

So

$$d_5=R^{10}{b}_5$$

Using the relation between $a,d_5,b_5$ ($a=5{ϵ}_0(R^4{b}_5+\frac{1}{R^6}{d}_5)$), we see that

$$b_5=\frac{a}{10{ϵ}_0{R}^4}$$

And because $d_5=R^{10}{b}_5$, we have that:

$$d_5=\frac{a{R}^6}{10{ϵ}_0}$$

We can now conclude with:

$$\begin{array}{rl}{V}_{\text{inside}}(s,\varphi )& =s^5\frac{a}{10{ϵ}_0{R}^4}\sin \left(5\varphi \right)\\ V_{\text{outside}}& =s^{-5}\frac{a{R}^6}{10{ϵ}_0}\sin \left(5\varphi \right)\end{array}$$

Here are some plots of the potential. A circle of radius $R$ is shown in dashed yellow. Notice how the inside solution (orange) joins the outside solution (cyan) continuously.

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