# Electric Potential/Energy question

1. Nov 23, 2003

### Flaming Toilet

Here's the question: "How much work must be done to bring three electrons from a great distance apart to within 1.0 x 10^-10 m from each other?"

Here's what I did :

.(work to bring one to another)

Work = PE + KE
= PE + 0
=q(V final - V initial) ; V initial = 0
=qV final
=q(KQ/r)
= (1.6 x 10^-19)((9 x 10^9 x 1.6 x 10^-19)/(1.0 x 10^-10))
= 2.3 x 10^-18 J

.(work to bring the other one)

Work = q(KQ/r)
= (1.6 x 10^-19)((9 x 10^9 x 1.6 x 10^-19)/(2 x 1.0 x 10^-10))
= 1.2 x 10^-18 J

. I then add them together to get the total work done

(2.3 x 10^-18 J) + (1.2 x 10^-18 J)
= 3.4 x 10^-18 J

Now, is that right? Feel free to correct me.

2. Nov 23, 2003

### Ambitwistor

I may be interpreting the question differently from you. I interpret it to mean that the three electrons are brought into the configuration of an equilateral triangle, with each edge of the triangle r=1 angstrom in length.

If so, then you can compute the final potential energy of one of the pairs of charges with ke2/r, and the final total potential energy of the three charges is three times that (energy from three pairs of charges).

3. Nov 23, 2003

### Flaming Toilet

I see. I'm visualizing the question as three electrons coming together in a line. Bleh, the book didn't provide a diagram. However, I think you're right. Thanks for guiding me in the right direction.