Question

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?

Answer 1

216.97 X 10⁻⁵ N

Step-by-step explanation:

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

F₁ =
`(9*10^9*8*7.7*10^(-18))/((2.6*10^(-2))^2)`

F₁ = 82.01 X 10⁻⁵ N

F₂=
`(9*10^9*8*15.4*10^(-18))/((2.6*10^(-2))^2)`

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