Calculating Potential from 3 Charges Placed on a Line

[kreil]

Homework Statement

Three charges +q, -2q, +q are placed on a line with equal separation a. Show that the potential at a point far away from the charges is approximately given by:

\Phi(r, \theta)=\frac{a^2q^2}{4 \pi \epsilon_0} \frac{3cos^2 \theta-1}{r^3}

Homework Equations

E-Field due to the discrete distribution of charges:
E(r)=\frac{1}{4 \pi \epsilon_0} \sum_{i=1}^3 q_i \frac{r-r_i}{|r-r_i|^3}

where r is the vector to the point where we want the field and r_i is the vector to charge q_i.

Potential:
\Phi(r)=-\int_O^rE \dot dr

The Attempt at a Solution

Position the charges such that the -2q is at the origin, and the +q charges are at +-a on the x -axis. Then the vectors to the charges are:

charge at +a: |r-a|=\sqrt{r^2+a^2-2racos\theta}=r(1-cos^2\theta)^{1/2} since a = r cos O.

charge at -a: |r+a|=\sqrt{4a^2+r^2+a^2+2racos\theta}=r(1+3cos^2\theta)^{1/2}

charge at origin: r=0.

So the field at r is:

E(r)=\frac{q}{4 \pi \epsilon_0} \left[ \frac{r-a}{|r-a|^3}+\frac{r+a}{|r+a|^3}-\frac{2r}{|r|^3} \right]

= \frac{q}{4 \pi \epsilon_0} \left[ \frac{r(1-cos^2 \theta)^{1/2}}{r^3(1-cos^2\theta)^{3/2}}+\frac{r(1+3cos^2\theta)^{1/2}}{r^3(1+3cos^2\theta)^{3/2}}-\frac{2}{r^2} \right]

= \frac{q}{4 \pi \epsilon_0} \frac{1}{r^2} \left[ \frac{1}{1-cos^2\theta}+\frac{1}{1+3cos^2\theta}-2 \right]

is this correct so far? After further simplifying this answer I obtained an equation for the field which seemingly does not produce a correct potential (I'm aware I will have to take the limit of the potential). Intuitively this field seems incorrect since the potential relies on 1/r^3 (i.e. the e-field should rely on 1/r^4 since the potential is the integrated e-field).

Thanks for any comments.

 

[kreil]

Ignoring the fact that i made many gross errors in the original post, the expression I'm trying to simplify now is:

E(\vec r)= \frac{q}{4 \pi \epsilon_0} \left[ \frac{\vec r + \vec a}{|\vec r + \vec a|^3}+\frac{\vec r - \vec a}{|\vec r - \vec a|^3}-\frac{2 \vec r}{|\vec r|^3} \right]

with the following:

|\vec r + \vec a| = \sqrt{r^2+a^2+2racos \theta}

|\vec r - \vec a| = \sqrt{r^2+a^2-2racos \theta}

I'm having a lot of trouble simplifying this because I'm out of practice with vectors, and I'm sure that it is not nearly as difficult as I am making it out to be (after 20 minutes of combining into one fraction and simplifying I decided it can't be the best method). Can someone give a hint?

E(\vec r)= \frac{q}{4 \pi \epsilon_0} \frac{1}{r^3} \left[ \frac{\vec r + \vec a}{(1+\frac{a^2}{r^2}+\frac{2a}{r}cos \theta)^{3/2}}+\frac{\vec r - \vec a}{(1+\frac{a^2}{r^2}-\frac{2a}{r}cos \theta)^{3/2}}-2 \vec r \right]
Thanks

 

[gabbagabbahey]

Why are you calculating the electric field first? Save yourself some trouble and just use the fact that the potential due to a point charge q located at \textbf{r}' is given by

\Phi=\frac{q}{4\pi\epsilon_0}\frac{1}{|\textbf{r}-\textbf{r}'|} (Taking infinity as the reference point)

This allows you to calculate the exact potential with ease. Then, simply use your favorite method for approximating a function (Taylor expansion perhaps) to see what the approximate potential is far from the origin.

 

[kreil]

you make a good point, thanks.

 

[kreil]

using that method i get a potential that relies on 1/r for r>>a..how in the world am i supposed to obtain a potential that relies on 1/r^3??

 

[vela]

The terms proportional to 1/r should cancel out. You probably made an algebraic error.

 

[kreil]

I am missing something substantial here.

\phi( \vec r) = \frac{q}{4 \pi \epsilon_o} \sum_{i=1}^N \frac{1}{|\vec r - \vec r_i|}

So for this charge dist.

\phi( \vec r) = \frac{q}{4 \pi \epsilon_o} \left[ \frac{1}{|\vec r - \vec a|} + \frac{1}{|\vec r + \vec a|}-\frac{2}{|\vec r|} \right]\phi( \vec r) = \frac{q}{4 \pi \epsilon_o} \left[ \frac{1}{\sqrt{r^2+a^2-2racos \theta}} + \frac{1}{\sqrt{r^2+a^2+2racos \theta}}-\frac{2}{r} \right]\phi( \vec r) = \frac{q}{4 \pi \epsilon_o} \frac{1}{r} \left[ \frac{(1+\frac{a^2}{r^2}-\frac{2a}{r} cos \theta)^{1/2}+(1+\frac{a^2}{r^2}+\frac{2a}{r} cos \theta)^{1/2}-(1+\frac{2a^2}{r^2}+\frac{a^4}{r^4}-\frac{4a^2}{r^2}cos^2 \theta)^{1/2}}{(1+\frac{2a^2}{r^2}+\frac{a^4}{r^4}-\frac{4a^2}{r^2} cos^2 \theta)^{1/2}} \right]

I must have been doing this part incorrectly. At this point, what is the best way to approximate these terms? Does a taylor expansion work even though we want r \rightarrow \infty after the approximation?

 

[vela]

r is large compared to a^2\pm2ra\cos\theta, so factor it out of the radical to get
\frac{1}{r\sqrt{1+[(a/r)^2\pm2(a/r)\cos\theta]}},

expand the square root as a Taylor series in powers of [(a/r)^2\pm2(a/r)\cos\theta], and collect terms in powers of r. This will give you a well-known result involving Legendre polynomials. If you already know what that result is, it can save you some time and quite a bit of algebra if you go straight to it.

 

[kreil]

that helped enormously, just finished the problem. thanks a ton!