Calculating Potential Difference for Electron Acceleration: A Step-by-Step Guide

[buttterfly41]

electricity problem :(

so here's the problem:

through what potential difference would an electron need to be accelerated for it to achieve a speed of 42.0% of the speed of light, starting from rest? The speed of light is c = 3.00e8 m/s

so i thought the equation i would use would be Vf-Vi=deltaPE / q

so i thougth the change in potential energy would be equal to the opposite of change in kinetic energy, so delta PE would = .5mv^2, and v is 3e8 X .42, so:

.5 X 9.11e-31 X (1.26e8)^2 / 1.6e-19 = 4520V, but this is wrong, so yea, no clue. any help would be much appreciated, thanks

 

[quantumdude]

You've got the right idea. All of the potential energy should show up as kinetic energy of the electron. Where you went wrong is right here:

so delta PE would = .5mv^2,

The expression $K=\frac{1}{2}mv^2$ is the nonrelativistic kinetic energy. In relativity there is a different expression:

$$K=\gamma\left(mc^2-1\right)$$
$$\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$$

If you use the correct expression, you should get the correct answer.

 

[m2003]

I would like to first commend you for attempting to solve this problem using the appropriate equations. However, there are a few things that need to be clarified. First, the equation you used, Vf-Vi=deltaPE/q, is actually the equation for electric potential, not potential difference. The correct equation for potential difference is V=Ed, where V is the potential difference, E is the electric field strength, and d is the distance traveled. Additionally, the equation you used for change in potential energy, deltaPE=.5mv^2, is correct for objects moving at non-relativistic speeds, but when dealing with particles approaching the speed of light, the correct equation is deltaPE=mc^2(1/sqrt(1-(v/c)^2)-1). Finally, the value you used for the velocity, 3e8 X .42, is incorrect. The correct way to calculate this would be to take 42% of the speed of light, so the velocity would be 0.42 X 3e8 = 1.26e8 m/s.

Using these corrections, the correct calculation would be:
deltaPE=9.11e-31 X (3e8)^2 (1/sqrt(1-(1.26e8/3e8)^2)-1) = 2.88e-11 J
Since potential difference is equal to change in potential energy divided by charge, we can rearrange the equation to solve for potential difference:
V=deltaPE/q = (2.88e-11 J)/(1.6e-19 C) = 1.8e8 V

Therefore, the potential difference required for the electron to achieve a speed of 42.0% of the speed of light is 1.8e8 volts. I hope this helps clarify the problem and solution for you. Keep up the good work in your scientific endeavors!