Calculating the work done to bring charges together

[NewtonianAlch]

Homework Statement

How much work is required to assemble eight identical point charges, each of magnitude q, at the corners of a cube of side s? Note: Assume a reference level of potential V = 0 at r = infinity. (Use k_e for ke and q and s as necessary.)

Homework Equations

V = E.s
U = q.V
V = kQ/s
U = kQ^2/s
F = kq^2/s^2

The Attempt at a Solution

In my first attempt I drew the cube, and the point charges, and realized this was essentially similar if you drew a sphere around the cube, so we could take the centre of the cube as the central point for all the charges.

So that distance (using pythagoras) would be x = 1/2*sqrt(3)*s

Then I decided the work done would simply be 8 times kq^2/(x)

i.e. (8k(q^2))/(1/2*sqrt(3)*s)

However this answer is wrong and I'm not sure why it would be wrong.

Looking for other possible ways, I was thinking calculating the work done to bring each charge individually to the vertex and doing a summation would be the solution, but this question is only worth 1 point so I'm thinking even though that could be right, this is not the way they want me to go ahead with this.

So then I though since the work done is equal to the integral of the force x change change in distance that this would amount to

U = 8 $\int$ kq^2/s^2 . ds

But I'm not too sure how to go about it with this, for starters there wouldn't be any force required to put the first charge in place since there's nothing else there repelling it, and then the change in distance would be from infinity to the vertex, I'm not too sure how that would be expressed. What should I be trying to do?

 

[SammyS]

The charges are all the same distance from the center of the cube. However, they are not all the same distance from each other.

Move the charges into place one at a time. How much work does it take to move first charge into place? Then how much for the second, then the third, etc.

 

[NewtonianAlch]

There is no work done to move the first charge into place because there is no prior electric field working against it, so that's 0.

To move the second charge, that would simply be kq^2/s

To move the third charge that would be kq^2/s + kq^2/sqrt(2)*s

...then the fourth, fifth, etc. then you add them altogether, is this what you mean?

 

[NewtonianAlch]

OK,

So after calculating the work done to move each charge into place against the electric fields present, I came to the answer:

12kq$^{2}$$\frac{1}{s}$ + 12kq$^{2}$$\frac{1}{\sqrt{2}s}$ + 4kq$^{2}$$\frac{1}{\sqrt{3}s}$

 

[ehild]

In case of interacting particles, you can calculate the potential energy by summing up the potential energies of all pairs, and then dividing by two, as every particle has been counted twice.
If you select a point charge of the cube, it has 3 neighbours at distance s, 3 other ones at distance √2s and one at distance√3s. The pairs (1,2), (1,3) ...(1,8) have the overall potential energy of

(kq²/s) (3+3/√2+1/√3)

The charges (2,1), (2,3) ... have the same contribution, but you see, that all pairs appear twice, so you have to multiply the contribution of one particle with N/2=4.

ehild

 

[NewtonianAlch]

In case of interacting particles, you can calculate the potential energy by summing up the potential energies of all pairs, and then dividing by two, as every particle has been counted twice.
If you select a point charge of the cube, it has 3 neighbours at distance s, 3 other ones at distance √2s and one at distance√3s. The pairs (1,2), (1,3) ...(1,8) have the overall potential energy of

(kq²/s) (3+3/√2+1/√3)

The charges (2,1), (2,3) ... have the same contribution, but you see, that all pairs appear twice, so you have to multiply the contribution of one particle with N/2=4.

ehild

What do you mean by multiply with N/2 = 4?

You mean I'd essentially have to divide my earlier equation by 2, and then multiply it by 4?

I can understand dividing by 2 to get rid of the double-counting scenario, but why multiply it by 4, because of the 4 particle interaction?

 

[gneill]

What do you mean by multiply with N/2 = 4?

You mean I'd essentially have to divide my earlier equation by 2, and then multiply it by 4?

I can understand dividing by 2 to get rid of the double-counting scenario, but why multiply it by 4, because of the 4 particle interaction?

You apply the formula for each of the vertexes of the cube (that is, you select each charge once and calculate the work to assemble all the pairs that you can make with that charge, then move on to the next charge, and so on). By symmetry, each of the sums so formed will be equal. That means you can apply the formula N = 8 times, or simply multiply the first instance by eight (simpler!).

When you're done you know that the method will have counted each pair twice. So you want to divide by two. The net result of multiplying the single instance of the formula by 8 and dividing by 2 is to simply multiply it by 8/2 = 4.