Question

A tiny spherical dust particle with mass 4.90×10−7 kg and charge 3.2 µC is released from rest a distance of 0.400 m above a large horizontal plastic sheet with uniform charge density 8.5×10−12 C/m2 . It arrives at a height 0.100 m above the sheet.

c. What is the speed of the particle at the position it arrives at?

Answer 1

given
mass, m = 4.90*10^-7 kg
charge, q = 3.2*10^-6 C
charge density of the sheet, sigma = 8.5*10^-12 C/m^2
h1 = 0.4 m
h2 = 0.1 m

electric field produced by the sheet, E = sigma/(2*epsilon)

= 8.5*10^-12/(2*8.854*10^-12)

= 0.480 N/C

Net force acting on the particle, Fnet = Fg - Fe

m*a = m*g - q*E

a = g - q*E/m

= 9.8 - 3.2*10^-6*0.28/(4.9*10^-7)

= 7.97 m/s^2

initial velocity, u = 0
let v is the velocity at height 0.100 m above the sheet.

use, v^2 - u^2 = 2*a*d

v^2 - 0^2 = 2*7.97*(0.4 - 0.1)

v = sqrt(2*7.97*0.3)

= 2.19 m/s <<<<<<<<<<<---------------------------Answer