An electron travels with speed 1.5 times 10^7 m/s between the two parallel charg

Question

An electron travels with speed 1.5 times 10^7 m/s between the two parallel charged plates shown in the figure below. The plates are separated by 1.0 cm and are charged by a 200 V battery. (a) What is the magnitude of the electric field between the plates? (b) What is the magnitude of the electric force on the electron? _______N What magnetic field strength and direction will allow the electron to pass between the plates without being deflected? (c) Magnitude of B = (d) Direction: (Remember that you want the direction of the magnetic force to be opposite to the direction of the electric force.)

Explanation / Answer

We know that:

Fe = q*E

Fe = 1.6*10^-19*20000

Fe = 3.2*10^-15 N

C.

Fe = Fm

qE = qVB

B = E/V

B = 20000/(1.5*10^7) = 1.33*10^-3 T

B = 1.33 mT

D.

Since direction of electric field is dowanward, So direction of magnetic field will be upward.

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