Voltage required to accelerate a proton into another one

In summary, to accelerate a proton towards another stationary proton that is "far away" outside of the proton gun and cause them to collide, the voltage across the parallel plates needs to be such that it will accelerate the proton to a kinetic energy of 1/2*QV. However, to achieve this, the equation U = kQq/r should be corrected to U = kQq/r^2, and the equation V = U/Q should not be used as it is not relevant to this scenario. Additionally, the distance r should be the sum of the charge radii of the two protons, rather than just one. The conservation of energy approach is still valid for solving this problem.
  • #1
aleksandrovich
2
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 
Last edited:
  • #5




You are correct in your observation that the proton would have to overcome an infinitely large potential energy to come in contact with the other one. This is because the Coulomb potential energy equation, U = kqQ/r, assumes that the two particles are point charges with no size or volume. In reality, protons have a finite size, so they cannot come into contact with each other.

To find the voltage required to accelerate the proton into the other one, we can use the conservation of energy equation, where the initial kinetic energy (KE) of the proton is equal to the final potential energy (PE) at the point of collision:

KE = PE

1/2 mv^2 = kqQ/r

We can rearrange this equation to solve for the voltage (V):

V = (2kqQ)/mr

Where m is the mass of the proton, r is the distance between the protons, q and Q are the charges of the protons, and k is the Coulomb constant.

However, as mentioned earlier, the protons cannot physically touch each other due to their finite size. Instead, they will reach a minimum distance where the repulsive force between them is equal to the attractive force, and they will stop accelerating towards each other. This minimum distance is called the Coulomb barrier and is dependent on the charge and mass of the particles.

So, in summary, the voltage required to accelerate a proton into another one would be given by the equation V = (2kqQ)/mr, but the protons will not physically touch each other due to their finite size and the presence of the Coulomb barrier.
 

Related to Voltage required to accelerate a proton into another one

1. What is the equation for calculating the voltage required to accelerate a proton into another one?

The equation for calculating the voltage required to accelerate a proton into another one is: V = (m1 + m2) * c^2 / q, where V is the voltage, m1 and m2 are the masses of the two protons, c is the speed of light, and q is the charge of the proton.

2. How does the voltage affect the acceleration of protons?

The voltage directly affects the acceleration of protons. The greater the voltage, the greater the acceleration of the protons will be.

3. Is the voltage required to accelerate a proton into another one dependent on the distance between the protons?

No, the voltage required to accelerate a proton into another one is not dependent on the distance between the protons. It is solely dependent on the masses and charges of the protons.

4. What is the unit of measurement for voltage in this context?

The unit of measurement for voltage in this context is volts (V). This is a unit of electric potential or potential difference, which is a measure of the energy required to move a unit of charge from one point to another.

5. How does the voltage required to accelerate a proton into another one compare to the voltage used in everyday electronics?

The voltage required to accelerate a proton into another one is much greater than the voltage used in everyday electronics. For example, a typical household outlet has a voltage of 120 volts, while the voltage required to accelerate protons is on the order of millions or even billions of volts.

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