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Interaction Energy For Two Point Charges

  1. Sep 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the interaction energy ( ##\epsilon_0 \int \vec{E_1}\cdot\vec{E_2}d\tau##) for two point charges, ##q_1## and ##q_2##, a distance ##a## apart. [Hint: put ##q_1## at the origin and ##q_2## on the z axis; use spherical coordinates, and do the ##r## integral first.]


    2. Relevant equations
    Interaction Energy is given by: $$ \epsilon_0 \int \vec{E_1}\cdot\vec{E_2}d\tau$$
    where ##d\tau## is the volume element, which in spherical coordinates is ##rsin\theta dr d\theta d\phi##

    3. The attempt at a solution
    So I know that $$\vec{E_1} = \frac{q_1}{4\pi\epsilon_0 r_1^2}\hat{r_1}$$ and $$\vec{E_2} = \frac{q_2}{4\pi\epsilon_0 r_2^2}\hat{r_2}$$
    So the dot product then is $$\vec{E_1}\cdot\vec{E_2} = \frac{1}{(4\pi\epsilon_0)^2} \frac{q_1}{r_1^2}\frac{q_2}{r_2^2}\hat{r_1}\cdot\hat{r_2}$$
    But since ##\hat{r}=\frac{\vec{r}}{|r|}##, this can be written as $$\vec{E_1}\cdot\vec{E_2} = \frac{1}{(4\pi\epsilon_0)^2} \frac{q_1}{r_1^3}\frac{q_2}{r_2^3}\vec{r_1}\cdot\vec{r_2}$$
    But I also know that the dot product can be expressed as ##\vec{E_1}\cdot\vec{E_2}=|E_1||E_2|cos\alpha## where ##\alpha## is the angle between them. So then ##\vec{E_1}\cdot\vec{E_2}## can also be written as $$\vec{E_1}\cdot\vec{E_2} = \frac{1}{(4\pi\epsilon_0)^2} \frac{q_1}{r_1^2}\frac{q_2}{r_2^2}cos\alpha$$
    So now, I'm not sure which version would be better to use because I don't know how to relate ##\alpha## to ##\theta## in the last version and for the first two versions, I'm not really sure how to define the position vectors because if ##q_1## is at the origin and ##q_2## is on the z axis then ##\vec{r_1}\cdot\vec{r_2}=0## and then the whole thing would just be 0 right? And when I go to integrate over ##r##, which ##r## is used in the volume element since there is ##r_1## and ##r_2##? I feel like there is some geometry that I am missing. And is the integration just the usual ##0\leq r \leq \infty##, ##0\leq\theta\leq \pi##, ##0\leq\phi\leq 2\pi##? Any help would be awesome!
     
  2. jcsd
  3. Sep 21, 2016 #2

    TSny

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    Why would ##\vec{r_1}\cdot\vec{r_2}=0##? (The two vectors are not the position vectors of the two particles.)
     
  4. Sep 21, 2016 #3
    Well my thinking was that if ##q_1## is at the origin then ##\vec{r_1}=(0,0,0)## and if ##q_2## is on the z axis then ##\vec{r_2}=(0,0,a)##. So then ##\vec{r_1}\cdot\vec{r_2}=r_{x1}r_{x2}+r_{y1}r_{y2}+r_{z1}r_{z2}=(0)(0)+(0)(0)+(0)(a)=0## right?
     
  5. Sep 21, 2016 #4

    TSny

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    ##\vec{r}_1## is not the position vector that locates ##q_1##.

    In the formula ## \vec{E} = \frac{kq }{r^3} \vec{r} ##, how would you describe the meaning of the vector ##\vec{r}##?
     
  6. Sep 21, 2016 #5
    Omg I think I got it. So ##\vec{r}## would be the vector describing the distance from the point charge to the reference point right? So then at some reference point P, the vector from ##q_1## would be ##\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}## and from ##q_2## it would be ##\vec{r}=x\hat{x}+y\hat{y}+(z-a)\hat{z}##. Is that right? so then $$E_1=kq_1\frac{(x\hat{x}+y\hat{y}+z\hat{z})}{(x^2+y^2+z^2)^{\frac{3}{2}}}$$ and $$E_2=kq_2\frac{(x\hat{x}+y\hat{y}+(z-a)\hat{z})}{(x^2+y^2+(z-a)^2)^{\frac{3}{2}}}$$. Is that correct?
     
  7. Sep 21, 2016 #6

    TSny

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    Yes.
     
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