Problem 22 - Arihant AIEEE Physics

Question

Two balls of same radius and mass are suspended on threads of length \(1 \mathrm{~m}\) as shown. The mass of each ball and charge is \(15 \mathrm{~g}\) and \(126 \mu \mathrm{C}\) respectively. When the balls are in equilibrium, the separation between them is \(8 \mathrm{~cm}\). The new saparation between them when one of the balls is discharged to half of original charge, is: (a) \(5 \mathrm{~cm}\) (b) \(6 \mathrm{~cm}\) (c) \(4 \mathrm{~cm}\) (d) \(2 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The new separation is 6 cm (option b).

Step 1: Understand the problem

We have a system with two identical charged balls, and they are suspended on threads. Initially, the equilibrium separation between them is given. We need to calculate the new separation when the charge on one of the balls is halved.

Step 2: Apply Coulomb's Law

Initially, the force of repulsion between the two charged balls can be given by Coulomb's Law:\[ F = k \frac{q^2}{d_0^2} \]where \( k \) is Coulomb's constant, \( q \) is the charge on each ball, and \( d_0 = 0.08 \) m is the initial separation.

Step 3: Consider Tension and Weight

In equilibrium, the tension in the thread also has a horizontal component balancing the electrostatic force and a vertical component balancing the weight of the balls. The forces acting horizontally give:\[ T\sin(\theta) = F \]where \( T \) is the tension, \( \theta \) is the angle with the vertical, and the vertical forces give:\[ T\cos(\theta) = mg \]where \( m = 0.015 \) kg and \( g = 9.8 \, \mathrm{m/s^2} \).

Step 4: Relate Angle to Separation

The relationship between the initial separation \( d_0 \) and \( \theta \) is:\[ \sin(\theta) = \frac{d_0}{2L} \]where \( L = 1 \) m is the length of the thread.

Step 5: Find New Equilibrium Position

When one charge is halved, let the new separation be \( d \). The electrostatic force becomes:\[ F' = k \frac{q \cdot \left( \frac{q}{2} \right)}{d^2} \]Given the same balance of forces in equilibrium, we set:\[ \frac{F'}{F} = \left( \frac{d_0}{d} \right)^2 = \frac{1}{2} \]

Step 6: Calculate the New Separation

Solving for \( d \), we find:\[ d = d_0 \sqrt{2} = 0.08 \sqrt{2} \approx 0.0566 \mathrm{~m} \]Converting to cm, \( d \approx 5.66 \mathrm{~cm} \), rounding options to the nearest one gives \( d = 6 \mathrm{~cm} \).

Key Concepts

Electrostatic Force

Electrostatic force is a fundamental concept in physics that describes the force between electrically charged objects. It's central to understanding how objects like the two balls in our exercise interact. Coulomb's Law is pivotal here. This law articulates that the electrostatic force (\( F \)) between two point charges is directly proportional to the product of the absolute values of the charges (\( q_1 \) and \( q_2 \)) and inversely proportional to the square of the distance (\( r \)) between them. The formula is given as:\[ F = k \frac{q_1 q_2}{r^2} \]Where \( k \) is Coulomb's constant (\( 8.9875 \times 10^9 \, \mathrm{N\,m^2/C^2} \)). In this exercise, each ball has the same initial charge, so the formula simplifies to an expression with one charge squared divided by the separation squared. This force is responsible for pushing the charged balls apart until the forces are balanced.

Equilibrium in Physics

In the context of physics, equilibrium refers to a state where all the forces acting on a system are balanced. For the system of the two charged balls, equilibrium is achieved when the combined effects of electrostatic repulsion and the gravitational force through the tension in the threads are balanced.There are two components of force acting on each ball:

• Horizontal: Caused by electrostatic force, expressed as \( T\sin(\theta) = F \) where the tension's horizontal component (\( T\sin(\theta) \)) balances the electrostatic force.
• Vertical: This involves gravitational force, expressed as \( T\cos(\theta) = mg \) which means the tension's vertical component counteracts gravity.

When the balls are at equilibrium, these components ensure that the forces don't cause any further movement, and the separation stays constant until other changes, like charge adjustment, are introduced.

Charge Distribution

Charge distribution is the arrangement of electric charge within a body. It determines how electric fields and forces manifest. In this problem, both balls initially have the same charge, creating a symmetrical force of repulsion. When one ball's charge is cut to half, the distribution is altered, making the understanding of force dynamics crucial.The change in charge impacts the electrostatic force magnitude according to Coulomb's Law, as now one charge becomes half. The effect on separation distance is visible from the modified formula: \[ F' = k \frac{q \cdot \left(\frac{q}{2}\right)}{d^2} \]This change requires recalculating the balance of forces as discussed in the equilibrium section. The resulting shift reduces the charge on one ball, therefore, affects the force, reducing the separation until it stabilizes. This becomes evident when recalculating the equilibrium distance with the equation \( \frac{d_0}{d} = \sqrt{2} \), revealing important insights into how charge distribution influences physical configurations.