(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: EM question

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