Potential difference of an electron

[zbtzylong]

Homework Statement

Through what potential difference would an electron need to be accelerated for it to achieve a speed of 46.0% of the speed of light, starting from rest? The speed of light is
c= 3.00 x 10^8 m/s
m_e = 9.11 x 10^-31 kg
|e| = 1.60 x 10^-19 C
Round your answer to three significant figures in units kV.

Homework Equations

V= Ed

The Attempt at a Solution

1. I tried to break down E into E= F/q.
2. F=ma
3. V = (ma/q)d
4. V = mad/q
5. V = m(d/t^2)d/q
6. V = m(d^2/t^2)/q
7. V = m(d/t)^2/q
8. V = mv^2/q

I plugged in the numbers using this and didn't get the right answer. Am I just going at this all wrong?

 

[G01]

You are going about this problem in a convoluted way, and may be making some mistakes along the way. Try this:

Remember that the kinetic energy gained by an electron going through a potential difference, V, is equal to the potential energy lost:

$$K_{gained}= U_{lost}=qV$$

Now, how much energy would an electron need to gain to go the given speed?

 

[zbtzylong]

I just don't feel like I'm wrapping my head about this correctly.

Using your K=qV formula, I worked the other way around.

K=1/2mv^2

Thus,

1/2mv^2 = qV
V = (mv^2)/(2q)

Using this, I still didn't get the correct answer.

 

[michalll]

Try using M = m/$$\sqrt{1 - (v/c)^2}$$

 

[G01]

I just don't feel like I'm wrapping my head about this correctly.

Try using M = m/$$\sqrt{1 - (v/c)^2}$$

This is probably your source of error. You need to use the relativistic mass of the electron, since you are moving at a significant fraction of the speed of light.

Nice catch michalll.