Question

Two identical charged spheres suspended from a common point by two massless strings of length \(\ell\) are initially a distance d \((d<<\ell)\) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the spheres approach each other with a velocity \(\mathrm{v}\). Then function of distance \(\mathrm{x}\) between them becomes \(\ldots \ldots\) (A) \(v \propto x\) (B) \(\mathrm{v} \propto \mathrm{x}^{(-1 / 2)}\) (C) \(\mathrm{v} \propto \mathrm{x}^{-1}\) (D) \(\mathrm{v} \propto \mathrm{x}^{(1 / 2)}\)

Short answer

The relationship between the velocity \(v\) and the distance \(x\) between the spheres is given by \(v \propto x^{-1}\). (Option C)

Step-by-step solution

So, the electrostatic force will be \(\text{F} = \frac{k \cdot q^2}{x^2}\). #Step 2: Determine the gravitational force acting on the spheres The gravitational force between two massive spheres is given by Newton's Law of Gravitation, which states that the force \(\text{F}\) = \(\frac{G \cdot m_1 \cdot m_2}{r^2}\), where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the spheres, and \(r\) is the distance between them. In this case, we can represent the masses as \(m\) and the distance as \(x\).

So, the gravitational force will be \(\text{F} = \frac{G \cdot m^2}{x^2}\). #Step 3: Find the effective force on the spheres The effective force on the spheres can be found by subtracting the gravitational force from the electrostatic force.

Thus, the effective force on the spheres is \(F_\text{eff} = \frac{k \cdot q^2}{x^2} - \frac{G \cdot m^2}{x^2}\). #Step 4: Determine the acceleration of the spheres using Newton's second law Newton's second law states that Force = mass × acceleration, i.e., \(\text{F} = \text{m} \cdot \text{a}\). In this case, the force acting on the spheres is \(F_\text{eff}\) and the mass is given by \(m\). We will denote the acceleration as "a" and divide the force by the mass to compute it.

So, the acceleration is \(\text{a} = \frac{F_\text{eff}}{m} = \frac{k \cdot q^2}{m \cdot x^2} - \frac{G \cdot m}{x^2}\). #Step 5: Find the relationship between the velocity \(v\) and the distance \(x\) The velocity function can be related to the acceleration using the integral \(\text{v}(\text{x}) = \int \text{a}(x) dx\). We can find the relationship by integrating the acceleration function with respect to \(x\).

First, we rewrite the acceleration function as \(\text{a} = A \cdot \frac{1}{x^2} - B \cdot \frac{1}{x^2}\), where \(A = \frac{k \cdot q^2}{m}\) and \(B = G \cdot m\).

Now, we integrate the function \(\int ( A \cdot \frac{1}{x^2} - B \cdot \frac{1}{x^2} ) dx\).

The result is \(\int ( A \cdot \frac{1}{x^2} - B \cdot \frac{1}{x^2} ) dx = A \cdot (-\frac{1}{x}) - B \cdot (-\frac{1}{x}) + C\), where \(C\) is the integration constant.

So, the relationship between velocity and distance is \(v(x) \propto \frac{1}{x}\), which corresponds to the option (C) \(v \propto x^{-1}\).

Key concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics, describing the force between two charges. The law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. This is expressed in the equation:

• \( F = \frac{k \cdot q_1 \cdot q_2}{r^2} \)

where:

• \( F \) is the electrostatic force.
• \( q_1 \) and \( q_2 \) are the charges.
• \( r \) is the distance between the charges.
• \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).

This law is crucial for understanding how charged objects interact with each other. In the exercise with charged spheres, the principle helps in calculating the repulsive force that causes the spheres to move apart initially. As charge leaks, the force diminishes, altering their motion dynamics.

Newton's Laws of Motion

Newton's Laws of Motion form the foundation of classical mechanics, detailing the relationship between a body and the forces acting upon it.

• The first law, or law of inertia, states that an object will remain at rest or in uniform motion unless acted upon by a force.
• The second law provides the relation \( F = m \cdot a \), where \( F \) is the force applied to an object, \( m \) is its mass, and \( a \) is its acceleration. This is key in determining the spheres' acceleration once forces are known.
• The third law states that for every action, there is an equal and opposite reaction.

In the scenario with charged spheres, the second law helps calculate acceleration. By equating the net force to mass times acceleration, one can find the dynamic movement of the spheres due to electrostatic and gravitational influences.

Gravitational Force

Gravitational force is the attraction between two masses. According to Newton's Law of Gravitation:

• \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \)

where:

• \( F \) is the gravitational force.
• \( m_1 \) and \( m_2 \) are the masses of the objects.
• \( r \) is the distance between the centers of the masses.
• \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \).

Despite being weak compared to electrostatic forces, gravity acts over larger distances and is vital in calculating the net force on the spheres. The gravitational force provides a counteracting influence on the spheres' separation, opposing the electrostatic force.

Integration in Physics

In physics, integration is often used to derive quantities from known gradients or rates. In our scenario, integrating acceleration with respect to distance helps in finding velocity as a function of position:

• Acceleration is determined by the effective force acting divided by the mass.
• The integral \( v(x) = \int a(x) \, dx \) illustrates how integrating acceleration over distance provides velocity.

For the charged spheres, we derive velocity as \( v(x) \propto \frac{1}{x} \) by integrating an expression derived from combined forces. This result assures us how velocity varies inversely with the distance between the spheres, insightfully capturing what happens as the spheres move closer and further apart.