Three charged particles form a triangle: particle 1 with charge Q1 = 80.0 nC is at xy coordinates (0, 3.00 mm), particle 2 with charge Q2 is at (0, −3.00 mm), and particle 3 with charge q = 18.0 nC is at (4.00 mm, 0). In unit-vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q2 is equal to (a) 80.0 nC and (b) −80.0 nC?

(a)  

(b)

Explanation:

For part (a)

Charge Q₁=+80×10⁻⁹C is on the y axis at y=0.003 m

And

Charge Q₂=+80×10⁻⁹C is on the y axis at y=-0.003 m

The force on particle 3 (which has charge of q=+18×10⁻⁹C) is due to vector sum of repulsive force from Q₁ and Q₂ as:

F₃₁+F₃₂=F₃

where

By using Pythagorean theorem r₃₁=r₃₂=0.005m

In magnitude angle notation the indicated vector addition becomes:

Therefor the net force is:

For Part (b)

Switching the sign of Q₂ amount to reversing the direction of its force on q.We have:

Therefor the net force is:

come closer and become far apart and become steam/gas

that would be because of both gravity and friction. so     c