# Electrostatics: Work required to assemble point charges

Tags:
1. Jun 13, 2015

### GreenLRan

I'm studying for the physics GRE and am fairly poor with EM.
1. The problem statement, all variables and given/known data

What is the work needed to assemble four point charges q into a regular tetrahedron of side length a?

2. Relevant equations
W = 1/2ΣqiV(ri)

3. The attempt at a solution

Assume that the origin is at one of the points, then the potential between each pair (there will be 3 pairs) is V = q/(4πεa). Since there are three pairs the V shows up 3 times in the summation. Also in the summation according to the work equation I wrote above you must multiple by q. Hence W = 1/2 * 3q2/(4πεa).

I get W = 1/2 * [ 3q2/(4πεa) ] , but the solutions says the work is 3q2 / (2πεa).

Why is there not a 2 * 4 *πεa in the denominator? Am I missing something?

Thanks!

Last edited: Jun 13, 2015
2. Jun 13, 2015

### Staff: Mentor

3. Jun 13, 2015

### GreenLRan

Assume that the origin is at one of the points, then the potential between each pair (there will be 3 pairs) is V = q/(4πεa). Since there are three pairs the V shows up 3 times in the summation. Also in the summation according to the work equation I wrote above you must multiple by q. Hence W = 1/2 * 3q2/(4πεa).

4. Jun 13, 2015

### theodoros.mihos

There is two ideas here. The potential and the potential energy. The potential is $\Sigma q_iV_i$ but the potential energy that we can get if destroy this distribution is the half. I think you have right.

5. Jun 13, 2015

### Staff: Mentor

There are three pairs for each charge. There are four charges (but avoid double-counting).

I don't see where your factor 1/2 comes from.

6. Jun 13, 2015

### theodoros.mihos

Ah. (3+2+1)/2 = 3. Se this

7. Jun 13, 2015

### GreenLRan

Do you mean you don't see where the factor 2 in the denominator of the solution's answer comes from? I don't see where that comes from either. I am assuming that my solution was correct then?

8. Jun 13, 2015

### theodoros.mihos

Ignore diferances of $a$. When the first go out the potential of the others is $3q/4\pi\epsilon_0a$. When the 2nd go out the potential of the others is $2q/4\pi\epsilon_0a$ and for the last dispartion the potential is $q/4\pi\epsilon_0a$. The total work is $(3+2+1)q^2/4\pi\epsilon_0a = 6q^2/4\pi\epsilon_0a$.

This is the reason for $1/2$ factor for charge distributions.

9. Jun 13, 2015

### Staff: Mentor

No. You introduce some factor 1/2 in post 3 without explaining it. Where does it come from?

10. Jun 13, 2015

### vela

Staff Emeritus
You've only considered one charge, the one you assumed is at the origin, interacting with the other three, so you're only taking into account the potential energy of the one charge. In other words, if you wanted to calculate the potential energy of the charge q at the origin, you'd have
$$U_q = \frac{1}{4\pi\epsilon_0} \frac{q q_1}{a} + \frac{1}{4\pi\epsilon_0} \frac{q q_2}{a} + \frac{1}{4\pi\epsilon_0} \frac{q q_3}{a},$$ where $q_i$ represent the other three charges. The first term represents the work required to bring $q$ in from infinity to overcome the repulsion due to charge $q_1$, and so on. There's no factor of 1/2, which, I assume, you grabbed from the formula you're trying to use. In this calculation, however, we're assuming the other three charges are already there when you brought charge $q$ in from infinity. To get the total potential energy in the system, we need to calculate the work required to assemble those three charges. Try this (not using the book's formula), and you'll see the book's answer is correct.

In terms of electric potential, the calculation above would be written
$$U_q = q\left[\frac{1}{4\pi\epsilon_0} \frac{q_1}{a} + \frac{1}{4\pi\epsilon_0} \frac{q_2}{a} + \frac{1}{4\pi\epsilon_0} \frac{q_3}{a} \right] = qV(\vec{r}_q).$$ The term in the square brackets is $V(\vec{r}_q)$. It's the electric potential at the point $\vec{r}_q$ where charge $q$ is. $U_q$ is one term in the formula you're trying to use from the book. You need additional terms corresponding to each of the other three charges. If you write this calculation out in gory detail, you should see why the factor of 1/2 appears in the book's formula if that's confusing you.

11. Jun 15, 2015

### GreenLRan

I see. Thank you. I believe I was interpreting the Work equation wrong. The 1/2 in the work equation only comes about when you count each potential twice (then you need to divide by 2). I was not counting each potential twice and dividing by 2.

Thanks!