Two very large parallel plates of charge are separated by 1.5 cm. The potential difference between the plates is 450 V. A small, positively charged particle flies horizontally between the plates with a speed of 3 x 10^5 m/s.
If we apply a uniform magnetic field, it is possible to allow the particle to pass through the plates without being deflected (i. e., the particle's velocity never changes direction).
(a) What is the direction of the magnetic field necessary to accomplish this goal?
(b) What is the magnitude of the magnetic field necessary to accomplish this goal?

Fm = Fe = 3*10^(4)*q

Explanation:

If we guest that the upper plate is the negative plate, the Electric field between the plates will make that the charge moves go up, that is, in the positive y-direction.

(a) The direction of the magnetic field must be in the negative y-direction.

(b) It is necessary that the magnetic field is equal to the electric field produced by the parallel plates. Hence, we can use the equation for the electric field E between parallel plates, and after that, we can compute the electric force:

E = V/d = 450/0.015 = 3*10^(4) N/C         (because 1.5cm = 0.015m)

And the force over the particle is

Fe = qE = 3*10^(4)*q

Finally the magnetic force must equal the electric force

Fm = Fe = 3*10^(4)*q

In this case is not necessary to use the value of the velocity

I hope this is usefull for you!

Regards,