A long non-conducting, massless rod of length L pivoted at its centre and balanced with a weight w at a distance x from the lefrt end. At the left and right ends fo the rod are attached small conducting sphers with positive charfges q and `2q` respectively. A distance h directly beneath each of these spheres is a fixed sphere with positive charge Q.
(a) Find the distance x where the rod is horizontal and balanced.b. What valueshold h have so that the rod exerts no vertical force on the bearing when the rod is horizontal and balanced?

Explanation

To solve the problem step by step, we will analyze the forces and torques acting on the rod and derive the necessary equations.

### Step 1: Understanding the Setup
We have a long non-conducting massless rod of length \( L \) pivoted at its center. A weight \( W \) is placed at a distance \( x \) from the left end. At the ends of the rod, there are small conducting spheres with charges \( q \) and \( 2q \) respectively. Below these spheres, at a distance \( h \), there are fixed spheres with charge \( Q \).

### Step 2: Analyzing the Torques
For the rod to be horizontal and balanced, the net torque about the pivot must be zero. The torques due to the weight \( W \) and the electrostatic forces from the charges must balance each other.

1. **Torque due to weight \( W \)**:
- The distance from the pivot to the weight is \( \frac{L}{2} - x \).
- Torque due to weight \( W \) is given by:
\[ \tau_W = W \left(\frac{L}{2} - x\right) \]

2. **Torque due to electrostatic forces**:
- The force between the charge \( q \) (left end) and charge \( Q \) (below it) is:
\[ F_1 = \frac{1}{4\pi \epsilon_0} \frac{qQ}{h^2} \]
- The torque due to this force about the pivot is:
\[ \tau_{F_1} = F_1 \left(\frac{L}{2}\right) = \frac{1}{4\pi \epsilon_0} \frac{qQ}{h^2} \cdot \frac{L}{2} \]

- The force between charge \( 2q \) (right end) and charge \( Q \) (below it) is:
\[ F_2 = \frac{1}{4\pi \epsilon_0} \frac{2qQ}{h^2} \]
- The torque due to this force about the pivot is:
\[ \tau_{F_2} = F_2 \left(\frac{L}{2}\right) = \frac{1}{4\pi \epsilon_0} \frac{2qQ}{h^2} \cdot \frac{L}{2} \]

### Step 3: Setting Up the Equation for Balance
Setting the total torque due to the electrostatic forces equal to the torque due to the weight:
\[ W \left(\frac{L}{2} - x\right) = \frac{L}{2} \left( \frac{1}{4\pi \epsilon_0} \frac{qQ}{h^2} + \frac{1}{4\pi \epsilon_0} \frac{2qQ}{h^2} \right) \]
Simplifying this gives:
\[ W \left(\frac{L}{2} - x\right) = \frac{L}{2} \cdot \frac{3qQ}{4\pi \epsilon_0 h^2} \]

### Step 4: Solving for \( x \)
Rearranging the equation to solve for \( x \):
\[ W \left(\frac{L}{2} - x\right) = \frac{3LqQ}{8\pi \epsilon_0 h^2} \]
\[ \frac{L}{2} - x = \frac{3qQ}{8\pi \epsilon_0 W h^2} \]
\[ x = \frac{L}{2} - \frac{3qQ}{8\pi \epsilon_0 W h^2} \]

### Step 5: Finding the Value of \( h \) for No Vertical Force
For the rod to exert no vertical force on the bearing, the upward electrostatic forces must equal the downward weight:
\[ W = \frac{3qQ}{4\pi \epsilon_0 h^2} \]
Rearranging gives:
\[ h^2 = \frac{3qQ}{4\pi \epsilon_0 W} \]
Taking the square root:
\[ h = \sqrt{\frac{3qQ}{4\pi \epsilon_0 W}} \]

### Final Answers
(a) The distance \( x \) where the rod is horizontal and balanced is:
\[ x = \frac{L}{2} - \frac{3qQ}{8\pi \epsilon_0 W h^2} \]

(b) The value of \( h \) such that the rod exerts no vertical force on the bearing is:
\[ h = \sqrt{\frac{3qQ}{4\pi \epsilon_0 W}} \]