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I have been given this question and would just like some hints for how to get started.

A system of charges consists of +2q at the origin and -q at the two points $(0, 0, a) $ and $(0, 0, -a)$. Show that the potential is given by:

$\phi(r, \theta) =\frac{q}{4\pi\epsilon_0}(\frac{2}{r}-\frac{1}{\sqrt{r^2+a^2-2racos\theta}}-\frac{1}{\sqrt{r^2+a^2+2racos\theta}})$

where r is the distance from the origin and $\theta$ is the polar angle of the field point. Any help is greatly appreciated.

dahaka5
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    Just add the potentials in cartesian coordinates and then transform to polar coordinates. – MrYouMath Oct 28 '17 at 13:44
  • So for example to see I'm on the right track the potential of the first charge +2q, it would be equal to $\frac{2q}{r4\pi\epsilon_0}$ ? – dahaka5 Oct 28 '17 at 13:57
  • Yes, that is your first component in the total potential. Just remember when you are setting up the other sources that you need to shift the origin. – MrYouMath Oct 28 '17 at 13:58
  • For the remaining two charges I have noticed that in the denominator of the solution it is using the cosine rule, but I do not know where that relates to? – dahaka5 Oct 28 '17 at 14:03
  • I suppose what I don't understand is how you get an angle between the charges when they are all in a straight line on the z-axis – dahaka5 Oct 28 '17 at 14:07

1 Answers1

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Hint:

if $\theta$ is the angle betwee the $z$ axis and the line passing thorough the origin and the point $P=(x,y,z)$ and $r$ is the distance $\overline{OP}$, than The distance from $P$ to the point $(0,0,a)$ is: $$ \sqrt{x^2+y^2+(z-a)^2}=\sqrt{x^2+y^2+z^2+a^2-2za}=\sqrt{r^2+a^2-2ar\cos \theta} $$ (because $z=r\cos \theta$).

Analogously you can find the distance from the oter charge in $(0,0,-a)$. Than use the fact that the potential is an additive function as noted in the comments.

Emilio Novati
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