A particle of charge + 13 µC and mass 3.57 10-5 kg is released from rest in a region where there is a constant electric field of + 495 N/C. What is the displacement of the particle after a time of 1.50 10-2 s?

20.3 mm

Explanation:

The force on the charge F = qE where q = charge = + 13 µC = + 13 * 10⁻⁶ C and E = electric field intensity = + 495 N/C.

The acceleration of the charge is thus a = F/m where m = mass of charge = 3.57 * 10⁻⁵ kg.

Its displacement is gotten from

s = ut + at²/2 where u = initial speed = 0 (since the charge starts from rest) and t = 1.5 * 10⁻² s.

s = 0 + at²/2 = at²/2

Substituting a and t into s we have

s = qEt²/2m = + 13 * 10⁻⁶ C * + 495 N/C * (1.5 * 10⁻² s) ² / (2 * 3.57 * 10⁻⁵ kg) = 0.0203 m = 20.3 mm