Three electrons form an equilateral triangle 1.00nm on each side. A proton is at the center of the triangle. What is the potential energy of this group of charges?

Explanation is in the file

cutt.ly/zxt1ipy

Ok let's work this out so We need to find the work done by the proton and the other two electrons on one electron in moving it in from infinity to its final location.. Then triple it.

I will call a side of the triangle s

You can put in the nanometer later.

I will call proton +q, electron -q

Now in final configuration how far is our electron from the proton?

xp = 2/3 altitude = (2/3) (s/2)sqrt 3)

I am just going to call it xp

now the work done on the electron moving from xp to pull it from there to infinity is its potential energy at infinity. In other words the PE is negative at xp if 0 at infinity for our problem with the proton (not so when we get to the electrons)

force = k (+q)(-q)/x^2

Pe =integral k q^2 dx/x^2 from xp to oo

= - k q^2 /xp

Now for the electrons (we have to push them together, compress the spring, so this PE will be +)

for each, final distance = s

so for each

PE = +k q^2/s and we have two of them

so

final PE = 2 k q^2/s - k q^2/xp

but xp = (2/3) s

so

PE = k q^2 ( 2/s - 1.5/s)

= (1/2)k q^2/s

what was the answer?