1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ You seem to be only off by a factor of 2. If you take the gradient of both of these functions, you get the same expression, except that one is twice the other. (Assuming that both of the results are written in spherical coordinates.) $\endgroup$
    – march
    Commented Feb 10, 2023 at 16:42
  • $\begingroup$ You were correct about the factor of two. I will take the gradient of both functions and get back to you. That might be it. TBD $\endgroup$ Commented Feb 10, 2023 at 18:05

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Beginning with "Another option is to start with..." is less clear than the first option. Please consider making a improvement. For example, what is $f$? $\endgroup$ Commented Feb 10, 2023 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.