Superposition of Electric Potential problem

[Lo.Lee.Ta.]

1. A system consists of the charges -q at (-d,0), +2q at (d,0), and +3q at (0,d). What is the total electric potential energy of the system.

2. I'm wondering, is it necessary to calculate each electric potential between charges from lower electron potential to higher electron potential?

-V_{-q to +3q} = $\frac{k(-q)}{\sqrt{2}d}$ + $\frac{k(+3q)}{\sqrt{2}d}$

V= -$\frac{2kq}{\sqrt{2}d}$



-V_{-q to +2q} = $\frac{k(-q)}{2d}$ + $\frac{k(+2q)}{2d}$

V= -$\frac{kq}{2d}$



-V_{+2q to +3q} = $\frac{k(+2q)}{\sqrt{2}d}$ + $\frac{k(+3q)}{\sqrt{2}d}$

V= -$\frac{5kq}{2d}$


The book solution says this last V should equal positive $\frac{k(2q)(3q)}{\sqrt{2}d}$...
But more simplified, that would equal: 6q²k/$\sqrt{2}$d

So, how did they get this? And- why is it positive?


I thought that if we have low electric potential to high electric potential, then ΔV= (-)

(V_a - V_b = ΔV = low - high = a NEGATIVE number)

In the first 2 calculations, we went from -q to +2q and -q to +3q.
So this means we went from LOW potential to HIGH potential. So it would make sense that these two V values are negative.

But the third also goes from lower to higher potential (+2q to +3q)... So why isn't this one also negative...?

#=_= So confused.

Please help? Thanks! :)

 

[Legaldose]

Okay, to find the electric potential due to two charges q and Q separated by a distance r, you need to use the equation:

V_q,Q = $\frac{kqQ}{r}$

To get the total electric potential you need to add up the electric potentials from each pair of charges. So for the last one you can see that it would in fact be what the book says. This is because both charges are positive. It doesn't matter in what order you calculate them, you will be adding them all up. So:

V_net = V_1,2 + V_1,3 + V_2,3

 

[Lo.Lee.Ta.]

Oh, wow!

My problem was not using the right formula! :P

Formula for electric potential energy between charges: U_q,Q = (kqQ)/r


Formula for electric potential between charges: V_q,Q = $\frac{kq}{x}$ + $\frac{kQ}{(x2)-x}$

Let me know if I have this right, please:
If I have 2 charges (q at x=5 and Q at x=20), and I want to know the electric potential between them, I would say:
Vq,Q=$\frac{kq}{5}$ + $\frac{kQ}{20-5}$

Is this the right expression?


Alright, back to my orginial problem...
Total electric potential energy of system:

U_total = $\frac{-kq(2q)}{2d}$ + $\frac{-kq(3q)}{\sqrt{2}d}$ + $\frac{k(2q)(3q)}{\sqrt{2}d}$

U_total = $\frac{3kq^2}{\sqrt{2}d}$ - $\frac{kq^2}{d}$

Thanks for the help! :)