Suppose an infinitely long cylindrical shell of radius $a$ carries a surface charge density $\sigma_0$ and is surrounded by a coaxial cable of inner radius $b$ and outer radius $c$ with uniform charge density $\rho_0$ such that the combined system is neutral.

I am asked to calculate the electrostatic energy in two equivalent ways, using $$W=\frac{1}{2}\int\rho V\,\mathrm{d}\tau$$ and $$W=\frac{\epsilon_0}{2}\int E^2\,\mathrm{d}\tau.$$ For some reason, which I have struggled (in vain) to locate, I am not obtaining the same result, and I would appreciate it if someone could help locate precisely what went wrong.

I think this is rather trivial, but it is tedious, and I must have overlooked something somewhere, embarrassingly enough.

First, I calculate $\rho_0$ in terms of $\sigma_0$. Consider some length $L$ along the system. The charge along the inner shell is $2\pi aL\sigma_0$ and the charge along the coaxial cable is $\pi c^2L\rho_0-\pi b^2L\rho_0=\pi L\rho_0(c^2-b^2)$. Setting their sum to be equal to 0, $$\rho_0=-\frac{2a\sigma_0}{c^2-b^2}$$

Next, I apply Gauss's law to find the electric field of the inner cylindrical shell and of the coaxial cable independently and sum them together to obtain the electric field for the system as a whole.

Consider first the inner cylindrical shell and construct a Gaussian cylinder of length L. Due to symmetry, the electric field must only point in the $\hat{\mathbf{s}}$ direction and is a function of $s$ only. For $s<a$, Gauss's law implies that the electric field is zero. For $s>a$, the charge enclosed is $Q=2\pi aL\sigma_0$, so $|\mathbf{E}|2\pi sL=Q/\epsilon_0$ yields $\mathbf{E}=(a\sigma_0/s\epsilon_0)\hat{\mathbf{s}}$.

Then consider the coaxial cable and a Gaussian cylinder of length L. Inside the cable, Gauss's law tells us the electric field is 0. For $b<s<c$, the enclosed charge is $Q=\rho_0\pi L(s^2-b^2)$, and the surface integral is equal to $|\mathbf{E}|2\pi sL$, so the electric field is $\mathbf{E}=(\rho_0/\epsilon_0)(s^2-b^2)(1/2s)\hat{\mathbf{s}}$, and similarly for $s>c$, it is $\mathbf{E}=(\rho_0/\epsilon_0)(c^2-b^2)(1/2s)\hat{\mathbf{s}}$.

Now, I combine these expressions to arrive at a piecewise function for the electric field.

$$ \mathbf{E}(\mathbf{r})=\left\{ \begin{array}{lr} 0 & s<a\\ \frac{a\sigma_0}{s\epsilon_0}\hat{\mathbf{s}} & a<s<b\\ \frac{a\sigma_0}{\epsilon_0}\frac{1}{s}\left(1+\frac{s^2-b^2}{b^2-c^2}\right)\hat{\mathbf{s}} & b<s<c\\ 0 & c<s \end{array} \right. $$

Now, from the electric field, we will calculate the potential, taking $\infty$ as our reference point and working our way inwards, to speak. The potential for $s>c$ is of course 0, because the electric field there is also 0.

Suppose that $b<s<c$. Then the potential is $V=V_1(s)$, where $$ \begin{align} V_1(s)&=-\int_c^s\left(\frac{a\sigma_0}{s'\epsilon_0}+\frac{\rho_0(s'^2-b^2)}{2s'\epsilon_0}\right)\,ds'=-\frac{a\sigma_0}{\epsilon_0}\int_c^s\frac{1}{s'}\left(1+\frac{s'^2-b^2}{b^2-c^2}\right)\,ds'=-\frac{a\sigma_0}{\epsilon_0}\left.\left(\frac{s'^2-2c^2\ln s}{2b^2-2c^2}\right)\right|_c^s\\ &=\frac{a\sigma_0}{2\epsilon_0(b^2-c^2)}\left(c^2-s^2+2c^2\ln(s/c)\right)=\frac{a\sigma_0}{\epsilon_0}\frac{1}{2}\left(\frac{c^2-s^2+2c^2\ln(s/c)}{b^2-c^2}\right) \end{align} $$

Suppose that $a<s<b$. Then the potential is $V=V_1(b)+V_2(s)$, where $$V_1(b)=\frac{a\sigma_0}{2\epsilon_0(b^2-c^2)}\left(c^2-b^2+c^2\ln(b/c)\right)=\frac{a\sigma_0}{\epsilon_0}\left(\frac{c^2\ln(b/c)}{b^2-c^2}-\frac{1}{2}\right)$$ $$ \begin{align} V_2(s)&=-\int_b^s\frac{a\sigma_0}{s'\epsilon_0}\,ds'=\frac{a\sigma_0}{\epsilon_0}\ln(b/s) \end{align} $$ Suppose that $s<a$. Then the potential is $V=V_1(b)+V_2(a)$, where $$V_2(a)=\frac{a\sigma_0}{\epsilon_0}\ln(b/a)$$

To summarize,

$$ V(s)=\left\{ \begin{array}{lr} \frac{a\sigma_0}{\epsilon_0}\left(\frac{c^2\ln(b/c)}{b^2-c^2}-\frac{1}{2}\right)+\frac{a\sigma_0}{\epsilon_0}\ln(b/a) & s<a\\ \frac{a\sigma_0}{\epsilon_0}\left(\frac{c^2\ln(b/c)}{b^2-c^2}-\frac{1}{2}\right)+\frac{a\sigma_0}{\epsilon_0}\ln(b/s) & a<s<b\\ \frac{a\sigma_0}{\epsilon_0}\frac{1}{2}\left(\frac{c^2-s^2+2c^2\ln(s/c)}{b^2-c^2}\right) & b<s<c\\ 0 & c<s \end{array} \right. $$

Next, I will proceed to calculate the electrostatic energy. Since I wish to obtain the energy per unit length, I will omit the integration along the $z$ axis from $z_0$ to $z_0+L$, because the multiplicative factor of $L$ will end up disappearing in the end anyway.

I can calculate the electrostatic energy with $W=\frac{1}{2}\int\rho V\,\mathrm{d}\tau$. We only need to consider the regions where $\rho$ is nonzero, namely the cylindrical surface and $b<s<c$. We have $$ \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\pi a\sigma_0\frac{a\sigma_0}{\epsilon_0}\left(\frac{c^2\ln(b/c)}{b^2-c^2}-\frac{1}{2}\right) + \pi a\sigma_0\frac{a\sigma_0}{\epsilon_0}\ln(b/a) \\ &\quad + 2\pi\frac{1}{2}\frac{-2a\sigma_0}{c^2-b^2}\int_b^c\frac{a\sigma_0}{2\epsilon_0(b^2-c^2)}s\left(c^2-s^2+c^2\ln(s/c)\right)\,ds\\ &=\frac{\pi a^2\sigma_0^2}{\epsilon_0}\left[\left(\frac{c^2\ln(b/c)}{b^2-c^2}-\frac{1}{2}\right)+\ln(b/a)+\frac{1}{(b^2-c^2)^2}\int_b^cs\left(c^2-s^2+2c^2\ln(s/c)\right)\,ds\right]\\ &=\frac{\pi a^2\sigma_0^2}{\epsilon_0}\left[\left(\frac{c^2\ln(b/c)}{b^2-c^2}-\frac{1}{2}\right)+\ln(b/a)+\frac{1}{(b^2-c^2)^2}\left.\left(c^2s^2\ln(s/c)-\frac{1}{4}s^4\right)\right|_b^c\right]\\ &=\frac{\pi a^2\sigma_0^2}{\epsilon_0}\left[\left(\frac{c^2\ln(b/c)}{b^2-c^2}-\frac{1}{2}\right)+\ln(b/a)+\frac{1}{(b^2-c^2)^2}\left(-b^2c^2\ln(b/c)-\frac{1}{4}c^4+\frac{1}{4}b^4\right)\right] \end{align} $$

Now, I will calculate the electrostatic energy with $W=\frac{\epsilon_0}{2}\int E^2\,\mathrm{d}\tau$. We have $$ \begin{align} W&=\frac{\epsilon_0}{2}\int E^2\,d\tau=\pi\epsilon_0\int_0^\infty E^2s\,ds=\pi\epsilon_0\frac{a^2\sigma_0^2}{\epsilon_0^2}\int_a^bs\frac{1}{s^2}\,ds + \pi\epsilon_0\frac{a^2\sigma_0^2}{\epsilon_0^2}\int_b^c\frac{1}{s}\left(1-\frac{s^2-b^2}{c^2-b^2}\right)^2\,ds\\ &=\frac{\pi a^2\sigma_0^2}{\epsilon_0}\left(\ln(b/a)+\frac{1}{(b^2-c^2)^2}\left.\left(c^4\ln s-c^2s^2+s^4/4\right)\right|_b^c\right)\\ &=\frac{\pi a^2\sigma_0^2}{\epsilon_0}\left(\ln(b/a)+\frac{c^4\ln c-c^4\ln b-c^4+b^2c^2+(1/4)c^4-(1/4)b^4}{(b^2-c^2)^2}\right) \end{align} $$

Upon examination, it seems they are not equal. Obviously, this is wrong.

Edit: I fixed it! They're equal now!

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    $\begingroup$ I have to upvote this just for the effort you put into asking and formatting this question! $\endgroup$ – Floris Apr 16 '14 at 1:36
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    $\begingroup$ There's an $s$ at the end of your last expression for $W$, which should be a $b$. Also, I think the expressions are the same - checking now. $\endgroup$ – Scott Lawrence Apr 16 '14 at 4:05
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    $\begingroup$ They're finally equal! Hooray! $\endgroup$ – Marcus Emilsson Apr 16 '14 at 4:17
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    $\begingroup$ Beat me by two minutes! $\endgroup$ – Scott Lawrence Apr 16 '14 at 4:19

You didn't really simplify the expressions in the same manner, so they're hard to compare. Even if they're not the same, it's often useful to have a clear idea of what differs in the final result. Simplifying the first expression to share the denominator $(b^2-c^2)^2$:

$$ \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\frac{\pi a^2\sigma_0^2}{\epsilon_0}\left[\left(\frac{c^2\ln(b/c)}{b^2-c^2}-\frac{1}{2}\right)+\ln(b/a)+\frac{1}{(b^2-c^2)^2}\left(-b^2c^2\ln(b/c)-\frac{1}{4}c^4+\frac{1}{4}b^4\right)\right]\\ &= \frac{\pi a^2\sigma_0^2}{\epsilon_0} \left[ \ln(b/a) + \frac{1}{(b^2-c^2)^2} \left(-\frac{1}{2}(b^4 + c^4) + b^2c^2 + b^2c^2\ln(b/c) + c^4\ln(b/c) -b^2c^2\ln(b/c) - \frac 1 4 c^4 + \frac 1 4 b^4 \right) \right]\\ \end{align} $$

Now the differences have been isolated to that numerator, which can be substantially simplified:

$$ \begin{align} &= -\frac{1}{2}(b^4 + c^4) + b^2c^2 + b^2c^2\ln(b/c) + c^4\ln(b/c) -b^2c^2\ln(b/c) - \frac 1 4 c^4 + \frac 1 4 b^4\\ &= b^2c^2 c^4\ln(b/c) - \frac 3 4 c^4 - \frac 1 4 b^4 \end{align} $$

Now simplifying the numerator of the second expression, compressing $\ln b - \ln c = \ln b/c$ and combining the $c^4$ terms, yields the same thing. So you did it right both ways.

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  • $\begingroup$ Thank you! I can't believe it took me so long to catch all my minor algebra mistakes--I edited it quite a few times! $\endgroup$ – Marcus Emilsson Apr 16 '14 at 4:31
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    $\begingroup$ I'm grateful to have reached the point where manipulations this complex are always done with the aide of a computer, if only as a check "am I allowed to make this transformation?". $\endgroup$ – Scott Lawrence Apr 16 '14 at 4:33

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