Shailesh K. answered 05/08/22
Expert Physics Tutor for Past 10 Years & MS in Electrical Engineering
Given: distances between charges d = 3.00 cm = 0.03 m.
The charges are q1 = +3e, q2 = -e, q3 = +e, and q4 = +8e, with e = 1.60 × 10-19 C.
Asked: What is the value of the net electrostatic force on (a) particle 1 and
(b) particle 2 due to the other particles?
Electrostatic force between two charges determined from Coulomb’s Law
F = ke qQ/d^2, q, Q charges, d- distance between them and
ke Coulomb constant = 8.988 x 10^9 kg m^2 s^-2C^-2
Net Electric force Q1 = Fq1q2 (right, attraction) + Fq1q3 (left, repulsion) + Fq1q4(left)
= (8.988 x 10^9 x 3 x 1.60 × 10^-19 x 1 x 1.60 × 10^-19)/ (.03)^2 –
(8.988 x 10^9 x 3 x 1.60 × 10^-19 x 1 x 1.60 × 10^-19)/ (.06)^2 –
(8.988 x 10^9 x 3 x 1.60 × 10^-19 x 8 x 1.60 × 10^-19)/ (.09)^2
= 7.67 x 10^-25 – 1.92 x 10^-25 – 6.82- x 10^-25 = -1.07 x 10^-25 N
Answer: Net force on particle 1: 1.07 x 10^-25 N left
Net Electric force Q2 = Fq2q1 (left, attraction) + Fq2q3 (right, attraction) + Fq2q4(right)
= -(8.988 x 10^9 x 3 x 1.60 × 10^-19 x 1 x 1.60 × 10^-19)/ (.03)^2 +
(8.988 x 10^9 x 1 x 1.60 × 10^-19 x 1 x 1.60 × 10^-19)/ (.03)^2 +
(8.988 x 10^9 x 1 x 1.60 × 10^-19 x 8 x 1.60 × 10^-19)/ (.06)^2
= -7.67 x 10^-25 + 2.56 x 10^-25 + 5.11 x 10^-25 = 0 N
Answer: Net force on particle 2: 0 N
I hope this helps.
Sincerely,
Shailesh (Sky) Kadakia, Expert Tutor WYZANT
Former Professor, University of Pittsburgh, Pittsburgh PA
Electrical Engineering Technology Department