# Electrostatic potential problem (got 1/3) correct

1. Oct 8, 2009

### hitman0097

Point charges q1, q2, and q3 are fixed at the vertices of an equilateral triangle whose sides are 2.50 m long. Find the electrostatic potential energy of this system of charges for the following charge values.
a.)q1=q2=q3=3.90uC
I got the right answer for this one 164mJ

b.)q1=q2=3.90uC,q3=-3.90uC
I used $$\Delta$$V=k[(3.90uC^2/2.5m)-3.90uC/2.5]=14080 (or something like that)
then V=q(sum of the charges) which is just 3.90uC * $$\Delta$$V
ans I got was 54.7mJ

c.)q1=q2=-3.90uC, q3=3.90uC
Same method as above.. same answer too.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 8, 2009
2. Oct 8, 2009

### rl.bhat

Hi hitman0097, welcome to PF.

3. Oct 9, 2009

### hitman0097

hmmm, I got the correct magnitude but the wrong sign? This is a online hw thing. But I don't know how you can minus energy as a answer. Like I know it's the change in energy and you can lose and gain it. But it can't ever be less then zero right?

4. Oct 9, 2009

### rl.bhat

In (c) there are two negative charges and one positive charge, So the net potential energy is negative. If you want to keep two positive charges at a certain distance, you have to push them towards each other.
If you want to keep one positive charges and one negative charge at a certain distance, you have to pull them apart from each other. First one you call it as positive PE and second one you call it as negative PE.