Solving Proton Acceleration & Velocity: Conservation of Energy

fubag
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[SOLVED] Conservation of Energy

Homework Statement



A proton (charge e, mass = 1.67 * 10^-27 kg) accelerates from rest by crossing a potential difference V = 3.00 V

a.) Find the velocity v_0 it acquires?

(Suppose this proton, having initial velocity v_0 from part (a), moves from very away directly towards a stationary heavy ion which is singly charged, i.e. having the same charge as the proton, +e.)


b.) Find the distance r_min between the proton and the ion at the moment of closest approach?

c.) Find velocity v of a proton when it is at a distance r = 2r_min from the ion

Homework Equations



a = (q/m)E

V = kQ/r

V = E*d (parallel plate)




The Attempt at a Solution



I am literally stuck for part a into determining what it could be, it seems something is missing in the problem.

Please help.

Thanks!
 
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fubag said:
I am literally stuck for part a into determining what it could be, it seems something is missing in the problem.

What is the work done on a charge q when it goes from a point with potential V1 to a point with potential V2?
 
work done is the change in potential energy so (V2 - V1)*q?

I was wondering could I just integrate the equation a = (q/m)E into v = (q/m)*V? because electric field is the derivative of potential difference?
 
fubag said:
work done is the change in potential energy so (V2 - V1)*q?

The change in KE is the work done. There's no need to integrate, since it is the potential difference which has been given.

I feel you should go through the basics once more by yourself.
 
I know change in potential energy = -Work

and that change in potential difference is the potential-energy change per unit charge...
so can't I just set Vq = U and then set that = (1/2)mv^2 and solve for v?
 
That's the correct way.
 

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