# Making a square with point charges (alternative solutions)

1. Sep 21, 2010

### narfarnst

1. The problem statement, all variables and given/known data
Calculate how much work it takes to make a square of side length A with point charges (all of charge q) at each vertex.

2. Relevant equations

3. The attempt at a solution
I know one way to do it. (find the difference in potential for each point and then go from there to find the work).
But I have two other ways that I'm not sure if they'll work or not.

First way
Since you'll have to do more work each time to overcome the newly added charges, can you just find the work it takes to place one charge and then multiply by six? (Because for each charge you have to do that much more work. i.e, 1 charge is 1W, 2 charges is 2W, 3 charges is 3W, so total charge would be (3*2*1)W=6W.... or would it be (3+2+1)W=6W?).

Second Way
Can I just find the potential at the center of the finished square and use that final potential to find work?

Also, am I correct in assuming that because there are initially no charges (and no surroundings to speak of) the first charges needs 0 work?

Thanks!!!

2. Sep 21, 2010

### kuruman

The best way to do this is to assemble the charges one at a time. The work to bring in the first charge is zero. For the second charge it is kq2/A. For the third charge it is qV, where V is the electric potential at the location of the third charge due to the presence of the other two charges, and so on. Add all the works to get the total. I wouldn't trust any other method.

You are correct in assuming that initially there are no charges. That's why it costs no work to bring in the first charge.

3. Sep 21, 2010

### fizzynoob

This is more of an accounting problem, no really a whole lot of math.

say you have charges oriented in this formation at the end
1 2

3 4

and they all have an equal distance of d

what is the energy to bring the first charge in?

1 -- all by itself

well its zero right. so U1 = 0

Now bring in charge 2

1 2

now we have work being done. -- the potential will be U2 = kq^2/d

now bring in charge 3

1 2

3 -- we are being affect by charge 1 and charge 2

U3 = Kq^2/d + kq^2/(d*sqrt(2))

Kq^2/d - being the potential due to charge 1 and 3
kq^2/(d*sqrt(2)) - being the potential due to charge 3 and 2, d*sqrt(2) is the distance between them

and lastly add the final charge

1 2

3 4

do you see the logic here?

once your done you sum up all of your energy

4. Sep 21, 2010

### narfarnst

Yes, I get that. That's the method I already know how to do (thanks, by the way. I hadn't actually worked it out yet, but I knew how :) ).
I'm asking if there is an easier way to do the problem, namely, one of the two ways I suggested.
Thanks.

5. Sep 21, 2010

### fizzynoob

>>Can I just find the potential at the center of the finished square and use that final potential to find work?

Oh alright :) well this is potential would be different in that scenario

that would be what the potential is at a point in the center of the square, not what work it took to build the square by pulling charges from infinity

the potential at the center of the square:

4*kq^2/(d*sqrt(2)/2) -- assuming all charges are the same

to build the square it would be :

(kq^2/d) * (4 + sqrt(2))

6. Sep 21, 2010

### narfarnst

A-ha!
Thanks.

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