- #1

Lo.Lee.Ta.

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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}= [itex]\frac{k(-q)}{\sqrt{2}d}[/itex] + [itex]\frac{k(+3q)}{\sqrt{2}d}[/itex]

V= -[itex]\frac{2kq}{\sqrt{2}d}[/itex]

-V

_{-q to +2q}= [itex]\frac{k(-q)}{2d}[/itex] + [itex]\frac{k(+2q)}{2d}[/itex]

V= -[itex]\frac{kq}{2d}[/itex]

-V

_{+2q to +3q}= [itex]\frac{k(+2q)}{\sqrt{2}d}[/itex] + [itex]\frac{k(+3q)}{\sqrt{2}d}[/itex]

V= -[itex]\frac{5kq}{2d}[/itex]

The book solution says this last V should equal

**positive**[itex]\frac{k(2q)(3q)}{\sqrt{2}d}[/itex]...

But more simplified, that would equal: 6q

^{2}k/[itex]\sqrt{2}[/itex]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! :)