Question

To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.7cm . Two of the particles have a negative charge: q1 = -7.7nC and q2 = -15.4nC . The remaining particle has a positive charge, q3 = 8.0nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Part B

Find the net force ?F? 3 acting on particle 3 due to the presence of the other two particles. Report you answer as a magnitude ?F3 and a direction ? measured from the positive x axis.

Express the magnitude in newtons and the direction in degrees to three significant figures.

Answer 1

(Image for representation purpose only)

Charge and their positions:

q₁ = -7.7 nC = -7.7 * 10^-9 C             at top vertex position.

q₂ = -15.4 nC = -15.4 * 10^-9 C         at right base vertex position.

q₃ = 8.0 nC = 8.0 * 10^-9 C               at left base vertex position.

r = 3.7 cm = 0.037 m

k = 8.99 * 10⁹ N.m² / C² is the Coulomb's force constant.

X component of the electric force on particle 3 due to particles 1 and 2 is,

F_x = k q₃ q₁ cos 60 / r² + k q₃ q₂ cos 0 / r²

F_x = 8.99 * 10⁹ * 8.0 * 10^-9 * 7.7 * 10^-9 * 0.5/ (0.037)² + 8.99 * 10⁹ * 8.0 * 10^-9 * 15.4 * 10^-9 * (1)/ (0.037)²

F_x = 1011.292 * 10^-6 N

Y component of the electric force on particle 3 due to particles 1 and 2 is,

F_y = k q₃ q₁ sin 60 / r² + k q₃ q₂ sin 0 / r²

F_y = 8.99 * 10⁹ * 8.0 * 10^-9 * 7.7 * 10^-9 * 0.866/ (0.037)² + 8.99 * 10⁹ * 8.0 * 10^-9 * 15.4 * 10^-9 * (0)/ (0.037)²

F_y = 350.311 * 10^-6 N

Magnitude of the net force on particle 3 is,

F = (F_x²+ F_y²)^1/2

F = 1070.247 * 10^-6 N

Direction of the electric field is

= tan^-1 (F_y / F_x)

= 19.106⁰