Question

2. An electron and a proton are each placed at rest in a uniform electric field of magnitude 4000 N/C. Determine the magnitude and direction of each particle's velocity 100.0 ns after being released.

Answer 1

The net force acting in the case above is the electric force.

Fnet = F(electric) = q E

ma = qE => a = qE/m

the acceleration of electron will be:

ae = 1.6 x 10^-19 x 4000/(9.1 x 10^-31) = 7.03 x 10^14 m/s^2

that for proton is:

ap =  1.6 x 10^-19 x 4000/(1.67 x 10^-27) = 3.83 x 10^11 m/s^2

time given is, t = 100 ns = 100 x 10^-9 s

we know from eqn of motion

v = u + at ; since started from rest, v = at

for electron, ve = ae t = 7.03 x 10^14 x 100 x 10^-9 = 7.03 x 10^7 m/s

for proton, vp = ap t = 3.83 x 10^11 x 100 x 10^-9 = 3.83 x 10^4 m/s

Hence, ve = 7.03 x 10^7 m/s and vp =  3.83 x 10^4 m/s

The direction of electron's will be the direction of electric field and proton will move in apposite direction. More information is required for direction.