find the net electrostatic force on the Q2 = 2.00−nC charged particle located at the origin. (Express your answer in vector form. Assume Q1 = −16.00 nC and Q3 = 5.00 nC.)

Question

Given:

Q1 has a point value of (-3.50cm, 0)

Q3 has a point value of (1.50cm, 0)

F = ? N

Please be detailed in your explanation so I can understand

Answer 1

Things to remember.

Forces are vectors so we need to take account of the direction (±x, ±y)

Electric forces act along the line between a pair of charges, like charges repel, unlike charges attract.

Consider each pair separately and find the vector sum.

Set q₂=2nC =2× 10^-9C (test charge), q₁=-16nC =-1.6×10^-8C (r₂₁=-3.5cm=.035m), q₃=5nC =5× 10^-9C (r₂₃=1.5cm=.015m)

The electric force between two charges distance r apart is given by

E=kq_aq_b/r² where k= 9 × 10⁹ and other quantities are given

(I dropped the subscripts for r but take note which one to use)

E₁= (9×10⁹ ×2×10^-9×1.6×10^-8)/(.035)² = 2.351×10^-4 N (attractive, so in the -x direction for q₂)

E₃= (9×10⁹×2×10^-9×5×10^-9)/(.015)² = 4×10^-4 N (repulsive, so in the -x direction for q₂)

E=E₁+E₃ =-6.351N