Potential due to line charge: Incorrect result using spherical coordinates

Question

Context

This is not a homework problem. Then answer to this problem is well known and can be found in [1]. The potential of a line of charge situated between $x=-a$ to $x=+a$ ``can be found by superposing the point charge potentials of infinitesmal charge elements. [1]'' Adjusting from [1] ($b\to a$), the answer to the problem below is $$ \boxed{ \Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \frac{Q}{2\, a} \left[ \ln{\left(\frac{ a+ \sqrt{a ^2 + r^2}}{ -a+ \sqrt{a ^2 + r^2}}\right)} \right] \, \,.} $$

Yet, because I am practicing using the curvilinear spherical coordinate system, I attempted to work this problem in that system. I know that $$\Phi( \mathbf{r} ) = \frac{1}{4\,\pi\,\epsilon_o} \int \frac { 1} { \left\| \mathbf{r}-\mathbf{r}^\prime\right\| }\rho(\mathbf{r}^\prime) \,d\tau^\prime \,$$ I also know that $$ \rho(r,\theta,\varphi) = \frac{Q}{2\,a} \,\frac{H{\left(r-0\right)}- H{\left(a-r \right)}}{1}\,\frac{\delta{\left(\theta-\frac{\pi}{2}\right)}}{r}\, \frac{\delta(\varphi-0) + \delta(\varphi-\pi) }{r\,\sin\theta} \,.$$ Further, since \begin{equation} \begin{aligned} x &= r \sin\theta \cos\varphi , \\ y &= r \sin\theta \sin\varphi , \\ z &= r \cos\theta , \end{aligned} \end{equation} I know that the expression of the distance between two vectors in spherical coordinates is given by the equation \begin{align} \|\mathbf{r}-\mathbf{r}^\prime\| = \sqrt{r^2+r'^2-2rr'\left[ \sin(\theta)\sin(\theta')\,\cos(\phi-\phi') +\cos(\theta)\cos(\theta')\right]}. \end{align} Finally, we are given that the obervation points, $\mathbf{r}$, are restricted as given by the equation $$\mathbf{r} = \left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right) .$$ Putting these togehter, we have that $$ \Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \int \frac { \frac{Q}{2\,a} \,\frac{H{\left(r^\prime-0\right)}- H{\left(a-r^\prime \right)}}{1}\,\frac{\delta{\left(\theta^\prime-\frac{\pi}{2}\right)}}{r^\prime}\, \frac{\delta(\varphi^\prime-0) + \delta(\varphi^\prime-\pi) }{r^\prime\,\sin\theta^\prime} } { \sqrt{r^2+r'^2-2rr'\left[ \sin(\theta)\sin(\theta')\,\cos(\phi-\phi') +\cos(\theta)\cos(\theta')\right]} } \, {r^\prime}^2\,\sin\theta^\prime\,dr^\prime\,d\theta^\prime\,d\phi^\prime \,.$$ Based on the point of observation, we rewrite the potential according to the equation $$ \Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \frac{Q}{2\,a} \, \int \frac { \left[H{\left(r^\prime-0\right)}- H{\left(a-r^\prime \right) } \right] \, \delta{\left(\theta^\prime-\frac{\pi}{2}\right)} \, \left[\delta(\varphi^\prime-0) + \delta(\varphi^\prime-\pi) \right] } { \sqrt{r^2+{r^\prime}^2-2\,r\,r^\prime\, \sin(\theta')\,\cos(\pi\pm \frac{\pi}{2}-\phi') } } \,dr^\prime\,d\theta^\prime\,d\phi^\prime \,.$$ Upon taking the angular integrals I rewrite the potential according to equation $$ \Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \frac{Q}{2\,a} \, \int_0^a \frac { 2 } { \sqrt{r^2+{r^\prime}^2 } } \,dr^\prime \,.$$ I know that $$ \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \ln{\left(x+ \sqrt{x^2 \pm a^2}\right)} \,. $$ Therefore, $$ \Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \frac{Q}{ a} \left[ \ln{\left(r^\prime+ \sqrt{{r^\prime}^2 + r^2}\right)} \right]_0^a \,. $$ Upon evaluation of the limits of integration, I have the incorrect result that $$ \boxed{ \Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \frac{Q}{ a} \left[ \ln{\left(\frac{ a+ \sqrt{a ^2 + r^2}}{r}\right)} \right] \, \,.} $$

Question

The result should be identical no matter what coordinate system that I choose. I have a gap in my understanding. Please help by identifying and stating the error in my analysis?

Bibliography

[1] http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potlin.html

Answer 1

Adjusting from [1] $(b→a)$, the answer to the problem below is $$\Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \frac{Q}{ a} \left[ \ln{\left(\frac{ a+ \sqrt{a ^2 + r^2}}{ -a+ \sqrt{a ^2 + r^2}}\right)} \right]$$

This is wrong by a factor of 2, in the original answer they use the linear density $\lambda$, which you substituted by $Q/a$, while it should be $Q/2a$.

$$\Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \frac{Q}{2\,a} \, \int_0^a \frac { 2 } { \sqrt{r^2+{r^\prime}^2 } } \,dr^\prime$$

From here you can use that the integrand is symmetric under the change of variables $r' \to -r'$, so you can write $2\int_0^a = \int_{-a}^a$ and the desired answer follows trivially.

Another option is to start with your final expresion $$\Phi{\left(r, \frac{\pi}{2},\pi\pm \frac{\pi}{2}\right)} = \frac{1}{4\,\pi\,\epsilon_o} \frac{Q}{ a} \left[ \ln{\left(\frac{ a+ \sqrt{a ^2 + r^2}}{r}\right)} \right]$$ And realise that $f(a):=\ln{\left(\frac{ a+ \sqrt{a ^2 + r^2}}{r}\right)}$ is odd under the change $a\to-a$. So you can rewrite $f(a)=\frac{1}{2}(f(a)-f(-a))$. This also gives you the desired result.