Question

In: Physics

To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three...

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.

Solutions

Expert Solution

(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,

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

Fx = 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

Fx = 1011.292 * 10-6 N

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

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

Fy = 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

Fy = 350.311 * 10-6 N

Magnitude of the net force on particle 3 is,

F = (Fx2+ Fy2)1/2

F = 1070.247 * 10-6 N

Direction of the electric field is

= tan-1 (Fy / Fx)

= 19.1060


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