Question

A first small bead of mass 2.2×10−2kg  and net charge 2×10−4C  is held at the origin, a second small bead of mass 1.6×10−2kg  and net charge 3.6×10−5C  is held at 0.16m  i^+  0m  j^ , and a third small bead of mass 6×10−2kg  and net charge −4.2×10−5C  is held at 0.16m  i^+  0.144m  j^ .

What is the x -component of the net electric force on the THIRD bead?

What is the y -component of the net electric force on the THIRD bead?

Asked this question once and the y-component was wrong. Would appreciate a correct answer.

Answer 1

let q1 = 2*10^-4 C
q2 = 3.6*10^-5 C
q3 = -4.2*10^-5 C

a) magnitude of force exerted by q1 on q3,

F1 = k*q1*q3/r13^2

= 9*10^9*2*10^-4*4.2*10^-5/(0.16^2 + 0.144^2)

= 1631 N (towards origin)

magnitude of force exerted by q2 on q3,

F2 = k*q2*q3/r13^2

= 9*10^9*3.6*10^-5*4.2*10^-5/(0.144^2)

= 656 N (towards -y axis)

so, Fnetx = F1x + F2x

= -1631*0.16/sqrt(0.16^2 + 0.144^2) + 0

= -1212 N <<<<<<<<<<<----------------------Answer

b) Fnety = F1y + F2y

= -1631*0.144/sqrt(0.16^2 + 0.144^2) - 656

= -1747 N <<<<<<<<<<<----------------------Answer