Understanding the Work-Energy Theorem for an Electron in a Circular Orbit

[Jaymo3141]

Homework Statement

(see attached)

Homework Equations

F=mv^2/r, Integral(F<dot>dl)

The Attempt at a Solution

the answer is B. Can anyone explain why?

 

[rude man]

Your F= mv^2/R is the force required to keep the elctron in its circular orbit. What generates that required force?

Hint #2: proton mass is >> electron mass so the proton effectively has very little v. But it has a huge m. Can we neglect the v of the proton in this problem, and if so, why?

 

[Jaymo3141]

well, I know that the force causing mv^2/r is kqq/r^2 coulombs law. And the proton is stationary so all the kenetic energy is from the electron. and I'm thinking that the energy of the electron can be calculated using work somehow since it is moving and there is a force on it. but the angle between the force and displacement is always 90, so Fdcos(theta) would always be 0.

 

[rude man]

Why don't you just equate the two forces & see what happens?

You're not exactly right when you say that the proton remains stationary. It also moves, and it has lots of mass so maybe Mv^2/2 of the proton is significant? You're right, it isn't, I just think you should be able to justify your assumption.

 

[rude man]

well, I'm thinking that the energy of the electron can be calculated using work somehow since it is moving and there is a force on it. but the angle between the force and displacement is always 90, so Fdcos(theta) would always be 0.

Since the k.e. ( 1/2 mv^2) of the electron is not changing, how can there be work done on it continuously? Its potential energy is not changing, and the work done would have to increase one or the other, right? So yes, the angle between the force and the displacement is always 90 deg.