Tension & Angle in Charge Suspension System in Space - Interactive Problem

Charged Spheres Suspended in Space

Two small spheres, each having the charge \(+Q\), are suspended by insulating threads of length \( L \) from a common hook. This arrangement is placed in space where there is no gravitational effect. Determine:

What are the angle between the two suspensions and the tension in each thread?

Select the correct option:

• A) \( 180^\circ, \quad T = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{(2L)^2} \)
• B) \( 90^\circ, \quad T = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{L^2} \)
• C) \( 180^\circ, \quad T = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{2L^2} \)
• D) \( 180^\circ, \quad T = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{L^2} \)

Detailed Explanation

In this problem, there is no gravitational force. The only forces acting on each sphere are:

• The tension \( T \) in the thread along the direction from the sphere to the hook.
• The repulsive electrical force \( F_e \) due to the other sphere.

For equilibrium, these two forces must balance along the line of action. Because there is no weight, the spheres will arrange themselves so that the threads lie along the line of the electrical force. This happens when the spheres are positioned directly opposite each other with respect to the hook, giving an angle of \( 180^\circ \) between the suspensions.

The electrical repulsive force between two charges is given by Coulomb's law: \[ F_e = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2}, \] where \( d \) is the distance between the centers of the spheres. Here, since both spheres hang from the same point and are exactly opposite, their separation is \( d = 2L \).

Substituting \( d = 2L \) into Coulomb's law gives: \[ F_e = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{(2L)^2}. \]

In equilibrium, the tension in the thread must balance this repulsive force, so: \[ T = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{(2L)^2}. \]

Therefore, the spheres will have an angle of \( 180^\circ \) between the suspensions and each thread will have a tension of \[ T = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{(2L)^2}. \]

This matches option A.