Voltage required to accelerate a proton into another one

AI Thread Summary
To determine the voltage required to accelerate a proton for fusion, the potential energy equations must be correctly applied. The initial equations presented were incorrect, particularly regarding the relationship between potential energy and distance. The charge radius of the proton should be considered to accurately calculate the distance when the protons are in contact. The conservation of energy approach is valid, but clarity on charge representation is necessary, as both protons share the same charge. Correcting these aspects will lead to a more accurate solution for the voltage needed.
aleksandrovich
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Homework Statement



A proton is accelerated through a voltage ∆V with parallel plates towards another stationary proton that is "far away" outside of the proton gun. To get fusion, we want them to get close enough to touch at the surfaces. Find the voltage across the plates that would accelerate the proton enough so that it will collide with the stationary proton.


Homework Equations



U = kqQ/r^2
U = 1/2*QV
Conservation of energy; KE = 1/2 mv^2

The Attempt at a Solution



Letting KE at the "front" of the proton gun be 0 and U at the end be 0, the voltage will accelerate the proton to KE=U=1/2*QV (similarly, let KE=0 at the point where the protons touch). The next step is where I got stuck; since touching the other stationary proton would require r to be zero, wouldn't the proton have to overcome an infinitely large potential energy to come in contact with the other one?
 
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aleksandrovich said:

Homework Statement



A proton is accelerated through a voltage ∆V with parallel plates towards another stationary proton that is "far away" outside of the proton gun. To get fusion, we want them to get close enough to touch at the surfaces. Find the voltage across the plates that would accelerate the proton enough so that it will collide with the stationary proton.

Homework Equations



U = kqQ/r^2
U = 1/2*QV
Conservation of energy; KE = 1/2 mv^2

The Attempt at a Solution



Letting KE at the "front" of the proton gun be 0 and U at the end be 0, the voltage will accelerate the proton to KE=U=1/2*QV (similarly, let KE=0 at the point where the protons touch). The next step is where I got stuck; since touching the other stationary proton would require r to be zero, wouldn't the proton have to overcome an infinitely large potential energy to come in contact with the other one?

The proton has a charge radius. Look this up (google and wiki should suffice).

But your first and second equations are wrong, and the third is irrelevant. Do you know where they're wrong?

Hints:

For the first equation, look at the power of r. What are the dimensions of the quantity in that equation?

For the second: Don't confuse this parallel plate setup with the potential energy acquired by a capacitor on being charged by Q to a voltage V. That's a different scenario.

For the third: you don't have to worry about speed here.
 
Sorry, I was being very sloppy. I should have said U = kQq/r and V = U/Q => U = QV.

Thanks for the tip about the charge radius, though. Using that for r, does the rest of the problem look good?
 
aleksandrovich said:
Sorry, I was being very sloppy. I should have said U = kQq/r and V = U/Q => U = QV.

Thanks for the tip about the charge radius, though. Using that for r, does the rest of the problem look good?

Is there a need to denote two different charges with Q and q? Isn't proton charge just +e?

You haven't shown exactly how you would work the problem, but apart from the issues discussed, the conservation of energy approach is sound.

EDIT: Just noticed what you were using for r. It's also off. When you have two identical balls touching, what's the distance between their centres?
 
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