In the figure below, particle 1 of charge +4 e is above a floor by distance d 1
ID: 2270580 • Letter: I
Question
In the figure below, particle 1 of charge +4e is above a floor by distance d1 = 1.40 mm and particle 2 of charge +6e is on the floor, at distance d2 = 6.20 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1?
In the figure below, particle 1 of charge +4e is above a floor by distance d1 = 1.40 mm and particle 2 of charge +6e is on the floor, at distance d2 = 6.20 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1?Explanation / Answer
q1 = +4e = +4*1.6*10^-19 C = 6.4*10^-19 C
q1 = +4e = +4*1.6*10^-19 C = 6.4*10^-19 C
q2 = +6e = +6*1.6*10^-19 C = 9.6*10^-19 C
r = sqrt(1.4^2+6.2^2) = 6.356 mm = 6.356*10^-3 m
F = k*q1*q2/r^2
= 9*10^9*6.4*10^-19*9.6*10^-19/(6.356*10^-3)^2
= 1.369*10^-22 N
theta = tan^-(1.4/6.2) = 12.72 degrees
Fx = F*cos(x) = 1.369*10^-22*cos(12.72) = 1.335*10^-22 N
q2 = +6e = +6*1.6*10^-19 C = 9.6*10^-19 C
r = sqrt(1.4^2+6.2^2) = 6.356 mm = 6.356*10^-3 m
F = k*q1*q2/r^2
= 9*10^9*6.4*10^-19*9.6*10^-19/(6.356*10^-3)^2
= 1.369*10^-22 N
theta = tan^-(1.4/6.2) = 12.72 degrees
Fx = F*cos(x) = 1.369*10^-22*cos(12.72) = 1.335*10^-22 N