Question

Particles of charge Q1 = +73 µC, Q2 = +44 µC, and Q3 = -80 µC are placed in a line. The center one is 0.35 m from each of the others.

Calculate the net force on each charge due to the other two. (State both magnitude and direction.

I only need help finding the net force on Q2. Please help if you can!

Answer 1

work out the force on a particle of charge Q = +1C sitting in the middle between the +73b and -80 charges.

use Couloumb's Law.


then, F = E q

Answer 2

I don't see how F = E q is needed here since there is no mention of an electric field in the question.

Answer 3

\[F=k|q_1q_2|/r^2\]\[F _{ q _{2}Net}= F_{1->2} + F_{3->2}\]The net force on two is the sum of the force from Q1 and Q3. These forces can each be calculated with the first equation, Coulomb's Law, by plugging in the constant k, the radius (given) and the respective charge (given). The signs of these charges will dictate which direction the resulting force will push. Q1 will repel Q2 (positive positive charge) and Q3 will attract Q2. This means both forces are in the same direction, which is along the positive x axis, defined as "to the right". That means both of the forces, the force on Q2 from Q1 and the force on Q2 from Q3, will be positive quantities and therefore sum to a positive force (to the right). Plug your values into a calculator and you should be good :)