How Is the Kinetic Energy of a Proton Calculated in an Electric Field?

[Peter G.]

Hi,

A proton initially at rest finds itself in a region of uniform electric field of magnitude 5.0 x 10⁶ Vm^-1. The electric field accelerates the proton for a distance of 1 km.

Find the kinetic energy of the proton.

So, what I did was the following:

KE = q * E * s

I then converted the result from J to MeV. I, however, get 5000 MeV and the book gets 500 MeV. Is my line of thought incorrect?

Thanks once again!

 

[vela]

Your answer is right for the given numbers.

 

[Peter G.]

Thanks! I checked and rechecked my calculations and numbers several times before posting here and I was about to go crazy! I guess the book made a mistake (everyone has the right to!)

 

[vela]

Given the numbers for the electric field and distance, you should be able to calculate the potential difference of 5000 MV between the two endpoints. Because the proton has charge e, to get the work done, you just stick an e in front of the V in the units. If you do that, you can see pretty easily 5000 MeV is correct. You don't need to worry that you made a mistake while converting units.

 

[Nickie]

Hello,

Your line of thought is not incorrect, but it seems like you may have missed a step in your calculation. The kinetic energy of a particle is given by the formula KE = 1/2 * mv^2, where m is the mass of the particle and v is its velocity. In this case, the proton is initially at rest, so its initial velocity is 0. Therefore, we can rewrite the formula as KE = 1/2 * m * (vf^2 - vi^2), where vf is the final velocity after being accelerated by the electric field and vi is the initial velocity (0). We can also use the formula for work, W = q * E * s, where q is the charge of the particle, E is the electric field, and s is the distance the particle travels. So, we can rewrite the kinetic energy formula as KE = q * E * s = 1/2 * m * (vf^2 - vi^2). Rearranging, we get vf^2 = (2qEs)/m. Now we just need to find the final velocity of the proton after traveling 1 km in the electric field. We can use the equations of motion to do this, specifically vf^2 - vi^2 = 2as, where a is the acceleration and s is the distance traveled. Since the proton is being accelerated by the electric field, we can substitute the electric field for the acceleration, giving us vf^2 = 2Es. Plugging this into our previous equation, we get KE = q * E * s = 1/2 * m * (2Es) = mEs. Now we just need to convert the result from J to MeV. The conversion factor is 1 MeV = 1.602 x 10^-13 J, so we can rewrite our formula as KE = mEs * (1 MeV/1.602 x 10^-13 J). Plugging in the values of m, E, and s, we get KE = (1.67 x 10^-27 kg) * (5.0 x 10^6 V/m) * (1000 m) * (1 MeV/1.602 x 10^-13 J) = 523 MeV, which is closer to the book's answer of 500 MeV. It's possible that you may have missed a conversion factor or made a