# Interaction Energy For Two Point Charges

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1. Sep 21, 2016

### thecourtholio

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. Sep 21, 2016

### TSny

Why would $\vec{r_1}\cdot\vec{r_2}=0$? (The two vectors are not the position vectors of the two particles.)

3. Sep 21, 2016

### thecourtholio

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?

4. Sep 21, 2016

### TSny

$\vec{r}_1$ 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 $\vec{r}$?

5. Sep 21, 2016

### thecourtholio

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?

6. Sep 21, 2016

Yes.