Charge to Mass Ratio of an Electron

Question

In what units must the radius (r) of the electron path be expressed when e/m is calculated? Why? And if the accelerating potential is doubled, what happens to r? Explain physically.

Thanks.

Answer 1

Any units will work fine, just make sure the units cancel in your final answer.  Putting everything in MKS usually works ok, so use meters.

I assume you're shooting an electron through a magnetic field perpendicular to the motion.  I will further assume it isn't so fast as to be relativistic.

So the kinetic energy is equal to the electrostatic potential:

1/2 mv^2 = qV

Moreover, the centripetal force is equal to the magnetic force:

mv^2 / r = qvB

Two equations in two unknowns--v and r.  Solve both equations for speed and set equal:

sqrt (2qV/m) = qBr/m

And solve for your radius:

r = sqrt (2Vm / qB^2)

So the radius goes like the square root of the potential.

If you wanted to solve for the charge to mass ratio:

q/m = 2V / (rB)^2

If V is in volts, r is in meters, and B is in tesla, q/m will come out in coulombs per kg.