Question

A very small object with mass 6.70×10⁻⁹ kg and positive charge 8.20×10⁻⁹ C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90×10⁻⁸ C/m2. The object is initially 0.450 m from the sheet.

What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.130 m?

Answer 1

given

m = 6.7*10^-9 kg
q = 8.2*10^-9 C
sigma = 5.9*10^-8 C/m^2
let d = 0.45 - 0.13 = 0.32 m
electric field prodcued by the sheet, E = sigma/(2*epsilon)

= 5.9*10^-8/(2*8.854*10^-12)

= 3332 N/C

use, Work-energy theorem

Work done by electric force = change in kinetic energy

Fe*d*(cos(180)) = KEf - KEi

q*E*d*(-1) = 0 - (1/2)*m*vi^2

-q*E*d = (1/2)*m*vi^2

==> vi = sqrt(2*q*E*d/m)

= sqrt(2*8.2*10^-9*3332*0.32/(6.7*10^-9))

= 51.1 m/s <<<<<<<<<-------------------Answer