# Homework Help: EM question

1. Feb 21, 2010

### kreil

1. The problem statement, all variables and given/known data
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:

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

2. Relevant 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:
$$\Phi(r)=-\int_O^rE \dot dr$$

3. 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).

2. Feb 23, 2010

### 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

Last edited: Feb 23, 2010
3. Feb 23, 2010

### 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.

4. Feb 23, 2010

### kreil

you make a good point, thanks.

5. Feb 23, 2010

### 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??

6. Feb 24, 2010

### vela

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

7. Feb 24, 2010

### 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?

8. Feb 24, 2010

### vela

Staff Emeritus
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.

9. Feb 24, 2010

### kreil

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