# Charge to mass ratio

1. May 30, 2009

### Linday12

1. The problem statement, all variables and given/known data
In an experiment similar to J. J. Thomson's, charged particles are observed to travel through a magnetic field of 0.040 T with a radius of curvature of 0.20 m. The path of these particles is made straight again when an electric field is introduced by two parallel plates.

V=200 V
distance between plates=10cm=0.10m
B=0.040T (assuming its parallel)

2. Relevant equations
r=mv/qB
ek=1/2(mv^2)=qV
v=$$\sqrt{2qV/m}$$
r$$^{2}$$=m2V/qB$$^{2}$$
q/m=2V/r$$^{2}$$B$$^{2}$$

3. The attempt at a solution
=2(200V)/(0.20m)$$^{2}$$(0.040T)$$^{2}$$
=6.3 x 10^6 C/kg

Does this look to be correct? I wasn't sure if I could do it all at once if the force was exerted at two different times (although it seems to make sense).

Thanks.

2. May 30, 2009

### LowlyPion

I don't think you can use kinetic energy here because it seems that the charge is traveling along an equipotential, i.e. not in the direction of the E field of the capacitor plates.

In the first case you found

mv2/R = qv*B

Then the E-Field is applied so there is no net force. This results in

qv*B = q*E

where E = V/d

v = E / B = V/(d*B)

If you apply this result to the first equation you should arrive at

q/m = V/(B2*R*d)

In this case since it happens that d = R it calculates to half of your result ...

q/m = V/(B*R)2

Last edited: May 30, 2009
3. May 30, 2009

### rl.bhat

In the relevant equation, the ek is the kinetic energy acquired be the charged particle before it enters the magnetic field. In the problem the given voltage is the voltage across the parallel plates which deflects the the beam of charged particles. In the calculation you have not used the distance between the plates which is given in the problem.

4. May 30, 2009

### Linday12

Wow, I can't believe I didn't notice that I hadn't used the plate separation. The problem makes complete sense now. Thank you both for your help, it was exactly what I needed!