Question

A -57.0 nC charged particle is released from rest in a uniform electric field of 0.141 N/C. The particle accelerates to a velocity of 0.0002c through a distance of 3.32 m. Determine the potential difference through which the particle accelerated, and determine the mass of the particle. Also, construct a diagram showing all significant aspects of the situation described (you should probably do this first).

Answer 1

The units N/C is the same as Volts/m and the distance is 3.32m so

potential difference = 0.141 ͯ*3.32 = 0.468 volts.

Let the mass of the particle be M.

Its acceleration in the field = [0.141 ͯ*57 ͯ*10^(-9)]/M = 8.04 ͯ*10^(-9)/M

Since it reaches a speed of 0.0002c (= 60000 m/s) after 3.32 m, its average speed over that distance = 30000 m/s

so the time taken = 3.32/30000. = 111*10^(-6) s. This means its (uniform) acceleration = 60000/(111*10^(-6)) = 542.2 ͯ*10^6 m/s, giving

M = 8.04 ͯ*10^(-9)/(542.2 ͯ*10^-6) = 14 82*10^(-18) kg