Question

A small particle has a mass of 3.5mg and carries a charge of 7.0μC. It is launched in the horizontal direction with a speed of 10.0m/s from an altitude of 0.25m above a large horizontal surface. There is a downward pointing electric field of strength 5.5N/C everywhere above the surface. How far does the particle travel in the horizontal direction before contacting the surface?

write up a neat solution showing all of the steps of your solution so that I can easily follow your reasoning.

Answer 1

Given,

m = 3.5 mg ; q = 7 uC ; v = 10 m/s ; h = 0.25 m ; E = 5.5 N/C

Fnet = Fe

ma = qE

a = qE/m = 7 x 10^-6 x 5.5/(3.5 x 10^-6) = 11 m/s^2

from eqn of motion

S = ut + 1/2 at^2

s = 0 + 1/2 a t^2

t = sqrt (2s/a) = sqrt (2 x 0.25/11) = 0.213 s

v = d/t => d = vt

d = 10 x 0.213 = 2.13 m

Hence, d = 2.13 m

------------------------------------

If we consider gravity,

a = 11 + 9.8 = 20.8 m/s^2

t = sqrt (2 x 0.25/20.8) = 0.155 s

d = 10 x 0.155 = 1.55 m

d = 1.55 m (considering gravity)