The figure shows two tiny 5.0-g spheres suspended from two very thin 1.0-m-long threads.The spheres repel each other after being charged to +91 nC and hang at rest as shown.What is the angle θ?
(k = 1/4πε₀ = 8.99 × 10⁹ N • m²/C²)

Question

Quizplus · Accepted Answer

Since the spheres are in equilibrium, the electrostatic force between them must be equal and opposite to the weight of each sphere. Thus, we have:
F = k(Q1Q2)/r^2 = mg
where k is Coulomb's constant, Q1 and Q2 are the charges on the spheres, r is the separation between the spheres, m is the mass of each sphere, and g is the acceleration due to gravity.

Solving for the separation between the spheres, we get:
r = √(kQ1Q2/mg)

Using the given values, we have:
r = √[(8.99 × 10^9 N•m^2/C^2)(91 × 10^-9 C)^2/(0.005 kg)(9.81 m/s^2)] = 0.1 m

Now, consider the triangle formed by the two threads and the vertical. Since the threads are very thin, we can assume they are parallel and the triangle is isosceles. Therefore, the angle between the threads is equal to half the angle between the threads and the vertical, which we can find using trigonometry:
tan(θ/2) = r/0.5 = 0.2
θ/2 = tan^-1(0.2) ≈ 4.1°
θ ≈ 8.2°

Therefore, the answer is A) 4.1° (rounded to one significant figure).

The Figure Shows Two Tiny 5