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Homework Help: EM question

  1. Feb 21, 2010 #1

    kreil

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    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:

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


    2. Relevant equations
    E-Field due to the discrete distribution of charges:
    [tex]E(r)=\frac{1}{4 \pi \epsilon_0} \sum_{i=1}^3 q_i \frac{r-r_i}{|r-r_i|^3}[/tex]

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

    Potential:
    [tex]\Phi(r)=-\int_O^rE \dot dr[/tex]


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

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

    charge at origin: r=0.

    So the field at r is:

    [tex]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] [/tex]

    [tex]= \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] [/tex]

    [tex]= \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] [/tex]

    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.
     
  2. jcsd
  3. Feb 23, 2010 #2

    kreil

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

    [tex]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] [/tex]

    with the following:

    [tex]|\vec r + \vec a| = \sqrt{r^2+a^2+2racos \theta}[/tex]

    [tex]|\vec r - \vec a| = \sqrt{r^2+a^2-2racos \theta}[/tex]

    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?

    [tex]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] [/tex]



    Thanks
     
    Last edited: Feb 23, 2010
  4. Feb 23, 2010 #3

    gabbagabbahey

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    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 [itex]q[/itex] located at [itex]\textbf{r}'[/itex] is given by

    [tex]\Phi=\frac{q}{4\pi\epsilon_0}\frac{1}{|\textbf{r}-\textbf{r}'|}[/tex] (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.
     
  5. Feb 23, 2010 #4

    kreil

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    you make a good point, thanks.
     
  6. Feb 23, 2010 #5

    kreil

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    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??
     
  7. Feb 24, 2010 #6

    vela

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    The terms proportional to 1/r should cancel out. You probably made an algebraic error.
     
  8. Feb 24, 2010 #7

    kreil

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    I am missing something substantial here.

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

    So for this charge dist.

    [tex]\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] [/tex]


    [tex]\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] [/tex]


    [tex]\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][/tex]

    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 [itex]r \rightarrow \infty[/itex] after the approximation?
     
  9. Feb 24, 2010 #8

    vela

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

    expand the square root as a Taylor series in powers of [itex][(a/r)^2\pm2(a/r)\cos\theta][/itex], 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.
     
  10. Feb 24, 2010 #9

    kreil

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