A particle of mass m = 1.6 x 10 -27 kg and charge q = +3.2 x 10 -19 C , moving w

Question

A particle of mass m = 1.6 x 10-27 kg and charge q = +3.2 x 10-19 C , moving with a constant horizontal velocity v = 4.0 x 106 m/s enters the constant electric field E = 20 x 102 N/C between the parallel plates of Fig. 9 below.

Find:

(a) the magnitude and direction of the electric force on the particle due to E

(b) the acceleration due to the electric field

(c) the acceleration due to the gravitational force

(d) how long it takes for the particle to leave the plates if the length of the plates is 0.80 m

(e) how far it will be from halfway between the plates as it leaves the plates.

Explanation / Answer

a) Force on the particle is given as F = qE and direction is in the direction of electric field magnitude of electric field = +3.2 x 10^-19 * 20 x 10^2 =64 x 10^-17 N and the direction of force is vertically downwards b) acceleration = force/ mass = 64 x 10^-17/1.6 x 10^-27 = 40 x10 ^ 10 m/s^2 c) acceleration due to gravity = mg /m= g = 9.8 m/s^2 d) time taken by particle to leave the plate = d/v = .8 / 4.0 x 10^6 = .2 * 10^ -6 sec e) vertical distance traveled = s s = u * t + 1/2*a *t^2 = 0 + .5 * 40 x10 ^ 10 * (.2 * 10^ -6)^2 = 2 millimeters

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