A proton is fired with a speed of 1.0e6 m/s through the parallel plate capacitor

Question

A proton is fired with a speed of 1.0e6 m/s through the parallel plate capacitor shown in figure. The capacitor's electric field is E vector = 1.0e5 (V/m), down. What magnetic field, B bar, both strength and direction, must be applied to allow the proton to pass through the capacitor with no change in speed or direction? Find the electric and magnetic fields in the proton's reference frame How does an experimenter in the proton's frame explain that the proton experience's no force as the charged plates fly by?

Explanation / Answer

a) F(magnetic) = F(electric)

q v B = q E

B = e/v

B = 1 x 10^5/1 x 10^6 = 0.1 T

Hence, B = 0.1 T

b)E' = E - v B

E' = 10^5 - 1 x 10^6 x 0.1 = 0 V/m

Hence, E' = 0 V/m

B' = B - VE/c^2

E = 0:

Hence, B' = 0.1 T

c)The proton is under the influence of two forces that balanes each other and let the proton pass through the plates undeflected. Hence the net force on the proton is zero in the experiment.

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