Question

A proton is located at the origin, and a second proton is located on the x-axis at x = 5.58 fm (1 fm = 10-15 m).
(a) Calculate the electric potential energy associated with this configuration.
1

(b) An alpha particle (charge = 2e, mass = 6.64 10-27 kg) is now placed at (x, y) = (2.79, 2.79) fm. Calculate the electric potential energy associated with this configuration.
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(c) Starting with the three particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place. (Throughout, neglect any radiation effects.)
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(d) Use conservation of energy to calculate the speed of the alpha particle at infinity.
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(e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity.
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Answer 1

No work shown

Answer 2

I just need an equation to get me started, the book does not list one that is useful. The example they give uses the equation V1 = Ke(q1/r1) the problem is I don't know q.

Answer 3

Look up the charge on a proton!!!!
(It is the same magnitude as the charge on an electron but positive, not negative, otherwise a hydrogen atom would not be neutral!!!)
Try U = k Q1 Q2 / r
where k = 9*10^9 N m^2/C^2

Answer 4

By the way an alpha particle is two protons.
Move alpha from infinity to the proper distance from first proton. Then double that for total gained by moving the alpha close to the two protons. Remember potentials can be added.