Final velocity of a proton through parallel plates

In summary, a moving proton with 6.4x10^-16 J of kinetic energy is accelerated by a potential difference of 5000 V between parallel plates. The proton emerges with a speed of approximately 1.3x10^6 m/s. This is found by using the equations q=1.60x10^-19 C, m=1.67x10^-27 kg, Ek=6.4x10^-16 J, and 5000V electric potential difference. By using the equations W=Vq and Ek=0.5mv^2, the answer is determined to be option C, 1.3x10^6 m/s.
  • #1
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Homework Statement



A moving proton has 6.4x10^-16 J of kinetic energy. The proton is accelerated by a potential difference of 5000 V between parallel plates.

http://members.shaw.ca/barry-barclay/Self-Tests/test09/elecst17.gif


The proton emerges from the parallel plates with what speed?

A. 8.8x10^5 m/s B. 9.8 x10^5 m/s C. 1.3x10^6 m/s D. 1.8x10^6 m/s

I know that the answer is C, I just don't know how to get there.

Homework Equations


q=1.60x10^-19 C
m=1.67x10^-27 kg
Ek=6.4x10^-16 J
5000V electric potential difference (I don't really fully grasp what this means. I have something that says electric potential difference = work done in moving test charge/magnitude of test charge [V=ΔE/q])


I know that Kinetic Energy=q|E|d
V=ΔE/q
W=Vq
Fe=kq/r2
|E|=V/d
Then I know the basic kinematics formulas

The Attempt at a Solution


I found work by using W=Vq. I get (5000V)(1.60x10-19C)= 8.00x10-16J

Then I'm lost as to what to do. But:

Using Ek=0.5mv2 I get the answer A. Is that actually the correct answer and the answer key from the thing he copied/pasted is wrong?

I saw this thread when googling https://www.physicsforums.com/showthread.php?t=168634 , which made me think that.
 
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  • #2
5000V of potential difference means for every coulomb of your charged particle, they will gain 5000J of energy so...

how much energy is gained/loss for a single particle?
how much energy does the particle end up with?
what speed does that translate to?
 

Related to Final velocity of a proton through parallel plates

1. What is the equation for calculating the final velocity of a proton through parallel plates?

The equation for calculating the final velocity of a proton through parallel plates is:
vf = (qEd)/(mp), where vf is the final velocity, q is the charge of the proton, E is the electric field strength, d is the distance between the plates, and mp is the mass of the proton.

2. How does the final velocity of a proton through parallel plates change if the electric field strength is increased?

If the electric field strength is increased, the final velocity of the proton will also increase. This is because the force on the proton is directly proportional to the electric field strength, according to the equation F = qE. Therefore, a higher electric field strength will result in a greater acceleration of the proton and a higher final velocity.

3. Can the final velocity of a proton through parallel plates be negative?

Yes, the final velocity of a proton can be negative. This means that the proton is moving in the opposite direction of the electric field. The direction of the final velocity is dependent on the direction of the electric field and the initial velocity of the proton.

4. What happens to the final velocity of a proton through parallel plates if the distance between the plates is increased?

If the distance between the plates is increased, the final velocity of the proton will decrease. This is because the electric field strength is inversely proportional to the distance between the plates, according to the equation E = V/d. Therefore, a greater distance between the plates will result in a weaker electric field and a lower final velocity for the proton.

5. How does the mass of the proton affect its final velocity through parallel plates?

The mass of the proton does not directly affect its final velocity through parallel plates. However, the mass is taken into account in the equation for final velocity, with a smaller mass resulting in a higher final velocity. This is because a lighter particle will experience a greater acceleration for the same force, according to Newton's second law of motion (F = ma). Therefore, a lighter proton will have a higher final velocity compared to a heavier proton under the same conditions.

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