Electrostatics: Work required to assemble point charges

[GreenLRan]

I'm studying for the physics GRE and am fairly poor with EM.
1. Homework Statement
What is the work needed to assemble four point charges q into a regular tetrahedron of side length a?

Homework Equations

W = 1/2Σq_iV(r_i)

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!

 

[mfb]

How did you get your answer?

 

[GreenLRan]

How did you get your answer?

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).

 

[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.

 

[mfb]

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.

 

[theodoros.mihos]

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

 

[GreenLRan]

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.

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?

 

[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.

 

[mfb]

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?

No. You introduce some factor 1/2 in post 3 without explaining it. Where does it come from?
The solution is right, your answer is not.

 

[vela]

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).

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.

 

[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!