Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.6 cm . two of the particles have a negative charge: q 1 = -7.7 nc and q 2 = -15.4 nc . the remaining particle has a positive charge, q 3 = 8.0 nc . what is the net electric force acting on particle 3 due to particle 1 and particle 2?

216.97 X 10⁻⁵ N

Explanation:

Charge q₁ and q₂ will attract q₃ with force F₁ and F₂ .F₁ and F₂ will be calculated as follows

F₁ =

F₁ = 82.01 X 10⁻⁵ N

F₂=

F₂ = 164.02 X 10⁻⁵ N

F₁ and F₂ will act at 60 degree so their resultant will be calculated as follows

R² = (82.01 X 10⁻⁵)² +( 164.02 X 10⁻⁵ )² + 2 X 82.01 X 164.02 X 10⁻¹⁰ Cos 60

R² = 47079.48 X 10⁻¹⁰

R = 216.97 X 10⁻⁵ N

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: