How Much Energy to Position Four Charges at Square Corners?

[flower76]

I'm not sure that I really understand this question:

How much energy is needed to place four positive charges, each of magnitude +5.0mC, at the vertices of a square of side 2.5cm?


what I was thinking is that V = kQ/r

And since all the charges are equal, and the same distance apart V1=V2=V3=V4 = (9x10^9)(5x10^-3C)/2.5x10^-2m = 1.8x10^9V

Vnet = V1 +V2 + V3 + V4 = 7.2x10^9V

However I haven't taken into account the diaganols - do I need to?

 

[Doc Al]

What you want to add is potential energy, not potential. The energy needed to bring two identical charges from infinity to a distance r apart is:
$$U = kQ^2/r$$

Hint: Imagine the four charges being brought from infinity, one at a time.

 

[flower76]

Ok I think I got it I added the potenial energy between each of the 6 charge sets and got 4.9x10^7J, hopefully that sounds about right.
I think I was getting things confused with the second part of the question:

Choose one way of assembling the charges and calculate the potential at each empty vertex as this set of charges is assembled. Clearly descrive the order of assembly.

Would I be correct to use V =kQ/r for this part of the question?

 

[Hootenanny]

Would I be correct to use V =kQ/r for this part of the question?

Yes, you should sum all of the individual potentials, for example;

$$V_{total} = \frac{kQ_{1}}{r_{1}} + ... + \frac{kQ_{n}}{r_{n}}$$

Where r_n is the distance from the charge Q_n to the empty vertex.

~H