Electrostatic potential problem (got 1/3) correct

[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 \DeltaV=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 * \DeltaV
ans I got was 54.7mJ

c.)q1=q2=-3.90uC, q3=3.90uC
Same method as above.. same answer too.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[rl.bhat]

Hi hitman0097, welcome to PF.
Your answers appear to be correct.

 

[hitman0097]

hmmm, I got the correct magnitude but the wrong sign? This is a online homework 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?

 

[rl.bhat]

hmmm, I got the correct magnitude but the wrong sign? This is a online homework 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?

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.