Question

In the diagram P1 and P2 are two parallel horizontal plates that are 5.4 mm apart in a vacuum and have a potential difference of 189 volts maintained between them, the upper plate being positive. Also, there is a horizontal magnetic field of 0.068. A horizontal beam of electrons is directed between the plates so that it is moving at right angles to the magnetic field as shown.

electorxxxxxx p2 x x x x x x x

a. Determine the magnitude of the force due to the electric field acting on each electron.

b. The direction of the magnetic force acting on the electron is

c. Calculate the speed of an electron that will allow it to pass undeflected between the plates. (Neglect the effect of gravity.)

Answer 1

(a) Magnitude of the potential difference between the two plates, P.D. = 189 Volt

Distance between the plates, d = 5.4 mm = 0.0054 m

So, magnitude of electric field between the two plates , E = P.D. / d

                                                                                                        = 189 / 0.0054 = 35000 V/m

Now, charge of an electron, q = 1.60 x 10^-19 C

Therefore, magnitude of force on each electron due to this electric field, F = q*E

                                                                                                                                      = (1.60 x 10^-19)*35000 N

                                                                                                                                      = 5.60 x 10^-15 N

(b) According to the right-hand rule, the direction of magnetic force will be in the downward direction.

(c) Suppose velocity of the electron to balance this force = v m/s

Therefore, magnetic force due to this velocity on the electron, Fm = B*q*v

Equalize this force with F –

Fm = F

=> B*q*v = q*E

Cancel ‘q’ from both sides –

=> B*v = E

=> v = E / B = 35000 / 0.068 = 5.15 x 10^5 m/s