To solve this problem, we need to analyze the forces acting on the two conducting balls before and after one of them is discharged. Initially, both balls are equally charged, and they repel each other due to the electrostatic force. When one ball is discharged, it loses its charge and will no longer experience the repulsive force from the other ball. Let's break this down step by step.
Initial Setup
We have two identical conducting balls, each with a charge \( Q \) and mass \( m \). They are suspended from a common point by long threads, and the distance between them is \( a = (108)^{1/3} \) cm. The equilibrium condition means that the gravitational force acting on each ball is balanced by the tension in the threads and the electrostatic repulsion between the two balls.
Forces in Equilibrium
In the initial state, the forces can be described as follows:
- Gravitational Force (Weight): \( F_g = mg \)
- Electrostatic Force: \( F_e = k \frac{Q^2}{a^2} \), where \( k \) is Coulomb's constant.
- Tension in the Thread: The tension can be resolved into vertical and horizontal components.
At equilibrium, the vertical component of the tension balances the weight, and the horizontal component balances the electrostatic force:
- Vertical: \( T \cos(\theta) = mg \)
- Horizontal: \( T \sin(\theta) = F_e \)
After Discharging One Ball
When one ball is discharged, it loses its charge and becomes neutral. The remaining charged ball will no longer exert a repulsive force on the discharged ball. Therefore, the only forces acting on the discharged ball are its weight and the tension in the thread.
New Equilibrium Condition
For the discharged ball, the equilibrium condition now only involves the weight and the tension:
- Vertical: \( T' \cos(\theta') = mg \)
- Horizontal: \( T' \sin(\theta') = 0 \) (since there is no electrostatic force).
As a result, the discharged ball will hang straight down, while the charged ball will remain at a distance \( b \) from the discharged ball. The distance \( b \) will be determined by the geometry of the setup.
Calculating the New Distance
Initially, the distance between the two balls was \( a \). After discharging one ball, the charged ball will still be at a height determined by the length of the thread \( L \). The discharged ball will hang directly downward. The new distance \( b \) can be visualized as follows:
Since the threads are long and the angle \( \theta' \) approaches 90 degrees for the discharged ball, the distance \( b \) will be equal to the length of the thread minus the vertical distance from the point of suspension to the discharged ball. However, since the threads are long and the angle is small, we can approximate the distance between the two balls as:
Given that the initial distance \( a \) is \( (108)^{1/3} \) cm, we can calculate:
Using the cube root, \( a \approx 4.78 \) cm. When one ball is discharged, the distance \( b \) will be approximately equal to the initial distance \( a \) since the charged ball will still remain at a distance close to \( a \) due to the long threads.
Final Result
Thus, the distance \( b \) between the two balls when equilibrium is restored is approximately:
b ≈ 4.78 cm
This analysis shows how the forces change when one of the charged bodies is neutralized, leading to a new equilibrium position based on the geometry of the system.
