Interaction Energy For Two Point Charges

[thecourtholio]

Homework Statement

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.]

Homework 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 $rsin\theta dr d\theta d\phi$

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!

 

[TSny]

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 $\vec{r_1}\cdot\vec{r_2}=0$

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

 

[thecourtholio]

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

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?

 

[TSny]

Well my thinking was that if $q_1$ is at the origin then $\vec{r_1}=(0,0,0)$

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

 

[thecourtholio]

No, $\vec{r}_1$ 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 $\vec{r}$?

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?

 

[TSny]

Yes.

 

[pathintegral]

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?

its correct but the integral you will do looks very very tedious and an elegant solution would be more welcome

 

[Steve4Physics]

its correct but the integral you will do looks very very tedious and an elegant solution would be more welcome

Note that the thread is 6+ years old!

 

[PhDeezNutz]

Interaction energy is merely potential energy of a system its nothing special.

U = kq1q2/r^2 will suffice.

That integral E^2 formula is only useful in cases of symmetry

I know this thread is old but whatevs.

 

[Orodruin]

Interaction energy is merely potential energy of a system its nothing special.

U = kq1q2/r^2 will suffice.

Supposedly the idea of the homework problem is to arrive at precisely this result starting from the interaction energy expression.

 

[PhDeezNutz]

Supposedly the idea of the homework problem is to arrive at precisely this result starting from the interaction energy expression.

Yeah just re-read the hint and that is indeed what the student is supposed to do.

Still a strange exercise considering that in Griffiths the E^2 result is derived from the more fundamental idea of summing over individual interactions using kqq/r^2. IIRC.

 

[minimikanix]

We have
$$\begin{align*}
\mathbf E_1 \cdot \mathbf E_2 &= \mathbf E_1 \cdot (-\nabla \cdot V_2).
\end{align*}$$
Also
$$\begin{align*}
\nabla \cdot (V_2 \mathbf E_1) &= \nabla V_2 \cdot \mathbf E_1 +
V_2 (\nabla \cdot \mathbf E_1) ,
\end{align*}$$
or
$$\begin{align*}
-\nabla V_2 \cdot \mathbf E_1 &= -\nabla \cdot (V_2 \mathbf E_1) +
V_2 (\nabla \cdot \mathbf E_1) .
\end{align*}$$
Similarly,
$$\begin{align*}
-\nabla V_1 \cdot \mathbf E_2 &= -\nabla \cdot (V_1 \mathbf E_2) +
V_1 (\nabla \cdot \mathbf E_2) .
\end{align*}
$$
Then,
$$
\begin{align*}
\epsilon_0\int\limits_{\text{all space}} \mathbf E_1 \cdot \mathbf E_2 \,d\tau
&= \frac{\epsilon_0}{2} \int \left[
\frac{\rho_1}{\epsilon_0}V_2 + \frac{\rho_2}{\epsilon_0}V_1
\right] \,d\tau\\
&= \frac{1}{2} \int
\left[
q_1 \delta (\mathbf r - \mathbf r_1) V_2(\mathbf r) +
q_2 \delta (\mathbf r - \mathbf r_2) V_1(\mathbf r)
\right] \,d\tau\\
&= \frac{1}{2}\left[
q_1 V_2(\mathbf r_1) + q_2 V_1(\mathbf r_2)
\right]\\
&=\frac{1}{2}\left[
q_1 \cdot \frac{k q_2}{a} + q_2 \cdot \frac{k q_1}{a}
\right]\\
&= \frac{k q_1 q_2}{a}
\end{align*}
$$
(where the other two terms were left out of the integral over space, as, after converting to surface integrals via the divergence theorem, we have that they both go to zero).