To solve the problem involving a non-conducting rod pivoted at its center with charges at its ends, we need to analyze the forces acting on the system and apply the principles of electrostatics and torque. Let's break this down step by step.
Understanding the Setup
We have a non-conducting rod of length \( l \) that is pivoted at its center. This means that the center of the rod is the point of rotation. The weight \( w \) is acting at a distance \( X \) from the left-hand side (L.H.S) of the rod. At each end of the rod, there are small conducting spheres with charges \( q \) and \( 2q \), respectively, positioned at a distance \( H \) from fixed charges \( Q \).
Identifying Forces and Torques
To find the distance \( X \), we need to consider the torques acting on the rod due to the weight and the electrostatic forces from the charges. The torque (\( \tau \)) is given by the formula:
Where \( r \) is the distance from the pivot to the point where the force is applied, and \( F \) is the force itself. In this case, we have:
- The gravitational force acting downwards at a distance \( X - \frac{l}{2} \) from the pivot.
- The electrostatic forces acting on the charges \( q \) and \( 2q \) at the ends of the rod.
Calculating Electrostatic Forces
The electrostatic force between two point charges is given by Coulomb's law:
- \( F = k \frac{|q_1 q_2|}{r^2} \)
In our case, the force between charge \( Q \) and charge \( q \) (at one end) and between charge \( Q \) and charge \( 2q \) (at the other end) can be calculated using their respective distances from the fixed charge \( Q \). The distances will be \( H \) for both charges since they are at the same height.
Setting Up the Torque Equation
Next, we set up the torque balance around the pivot point. The total torque due to the weight must equal the total torque due to the electrostatic forces for the system to be in equilibrium:
- Torque due to weight: \( \tau_w = w \cdot (X - \frac{l}{2}) \)
- Torque due to charge \( q \): \( \tau_q = F_q \cdot \frac{l}{2} \)
- Torque due to charge \( 2q \): \( \tau_{2q} = F_{2q} \cdot \frac{l}{2} \)
Setting the sum of torques equal to zero gives us:
- \( w \cdot (X - \frac{l}{2}) = F_q \cdot \frac{l}{2} + F_{2q} \cdot \frac{l}{2} \)
Solving for Distance \( X \)
Now, we can substitute the expressions for \( F_q \) and \( F_{2q} \) into the torque equation. After substituting and simplifying, we can isolate \( X \) to find its value:
- Combine the forces and simplify the equation.
- Rearrange to solve for \( X \).
Ultimately, the distance \( X \) can be expressed in terms of the known quantities \( w \), \( q \), \( 2q \), \( H \), and \( l \). This will yield a specific numerical value based on the parameters provided.
Final Thoughts
This problem illustrates the interplay between gravitational and electrostatic forces in a mechanical system. By carefully analyzing the torques and forces involved, we can derive meaningful relationships and solve for unknown distances. If you have specific values for \( w \), \( q \), \( 2q \), \( H \), and \( l \), we can plug those into our final equation to find the exact value of \( X \).
