# A problem with electric potential and conservation of energy

## Homework Statement

A) How much work would it take to push two protons very slowly from a separation of 2.00*10-10 m (a typical atomic distance) to 3.00*10-15 m (a typical nuclear distance)?

B) If the protons are both released from rest at the closer distance in part A, how fast are they moving when they reach their original separation?

## Homework Equations

Eq 1: Ka + Ua = Kb + Ub
Eq 2: Wa->b = Ua - Ub

## The Attempt at a Solution

Using Eq 1: to solve for vb:
0 + Ua - Ub = (1/2)mvb2
sqrt( 2(Ua - Ub)/m ) = vb

Where m = 1.67*10-27 kg, the mass of a proton.

From here, I plug in values and get vb = 9.58598*106 m/s.
Naturally, this is wrong.

I think the error is in that I worked the problem as though one of the protons is held still (which I'm sure is fine for part A, but apparently not B). If this is the case, how should I be handling this problem?

Edit: Whoops. Forgot to re-add this in when my first thread submission attempt got fried...
I've already calculated Wa->b = 7.6729*10-14 J. This, according to the homework problem, is correct.

Edit2: Epic fail. Accidentally posted in the wrong section. This should go under Introductory Physics, as it's just introducing calculus-based physics. If someone could move this thread to there, that would be appreciated.

Last edited:

## Answers and Replies

kuruman
Science Advisor
Homework Helper
Gold Member
When you release the protons in part B, both protons have kinetic energy. Therefore the change in potential energy is equally divided between the two protons. This is true because the masses of the two particles are the same. If they weren't, i.e. if you had a proton and a deuteron or alpha particle, you would have to conserve both energy and momentum to determine how the potential energy change is shared between the particles.

Thanks. I had the feeling it was something like that, but I'm always so insecure with physics, so I doubt my thoughts on it too frequently...

I've successfully solved the problem now. :)