A ball of mass `10^(-2) kg` and having charge `+ 3 xx 10^(-6)C` is tied at one end of a `1m` along thread. The other end of the thread is fixed and a charge `- 3 xx 10^(-6) C` is placed at this end. The ball can move in the circulr orbit of radius `1m` in the vertical plane. Initially, the ball is at the bottom. Find the minimum initial horizontal velocity of the ball, so that it will be able to complete the full circle.

Answer

To solve the problem, we need to find the minimum initial horizontal velocity of a ball with a mass of \(10^{-2} \, \text{kg}\) and charge \(+3 \times 10^{-6} \, \text{C}\) that is tied to a thread of length \(1 \, \text{m}\) and can move in a vertical circular motion around a charge of \(-3 \times 10^{-6} \, \text{C}\) located at the fixed end of the thread.

### Step 1: Understand the Forces Acting on the Ball
At the top of the circular path, the ball experiences two forces:
1. Gravitational force (\(F_g = mg\)) acting downward.
2. Electrostatic force (\(F_e\)) due to the charge at the fixed end, also acting downward.

The centripetal force required to keep the ball in circular motion at the top is provided by the sum of these two forces.

### Step 2: Write the Equation for Centripetal Force
At the top of the circle, the equation for centripetal force can be written as:
\[ F_g + F_e = \frac{mv'^2}{r} \]
Where:
- \(v'\) is the velocity of the ball at the top of the circle.
- \(r = 1 \, \text{m}\) is the radius of the circular path.

### Step 3: Calculate the Electrostatic Force
The electrostatic force \(F_e\) between the two charges can be calculated using Coulomb's law:
\[ F_e = k \frac{|q_1 q_2|}{r^2} \]
Where:
- \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\)
- \(q_1 = 3 \times 10^{-6} \, \text{C}\)
- \(q_2 = -3 \times 10^{-6} \, \text{C}\)
- \(r = 1 \, \text{m}\)

Substituting the values:
\[ F_e = 9 \times 10^9 \cdot \frac{(3 \times 10^{-6})(3 \times 10^{-6})}{1^2} = 9 \times 10^9 \cdot 9 \times 10^{-12} = 81 \times 10^{-3} \, \text{N} = 0.081 \, \text{N} \]

### Step 4: Set Up the Equation
Now substituting \(F_g\) and \(F_e\) into the centripetal force equation:
\[ mg + F_e = \frac{mv'^2}{r} \]
Substituting \(m = 10^{-2} \, \text{kg}\), \(g = 10 \, \text{m/s}^2\), and \(r = 1 \, \text{m}\):
\[ (10^{-2} \cdot 10) + 0.081 = \frac{10^{-2} v'^2}{1} \]
\[ 0.1 + 0.081 = 10^{-2} v'^2 \]
\[ 0.181 = 10^{-2} v'^2 \]
\[ v'^2 = 0.181 \times 10^2 = 18.1 \]
\[ v' = \sqrt{18.1} \approx 4.25 \, \text{m/s} \]

### Step 5: Apply Conservation of Energy
The total mechanical energy at the bottom must equal the total mechanical energy at the top:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mv'^2 + mg(2r) + F_e \cdot 2r \]
Where \(2r = 2 \, \text{m}\) since the ball moves from the bottom to the top.

Substituting the values:
\[ \frac{1}{2} (10^{-2}) v^2 = \frac{1}{2} (10^{-2}) (18.1) + (10^{-2})(10)(2) + 0.081 \cdot 2 \]
\[ \frac{1}{2} (10^{-2}) v^2 = 0.0905 + 0.2 + 0.162 \]
\[ \frac{1}{2} (10^{-2}) v^2 = 0.4525 \]
\[ v^2 = \frac{0.4525 \cdot 2}{10^{-2}} = 90.5 \]
\[ v = \sqrt{90.5} \approx 9.52 \, \text{m/s} \]

### Final Answer
The minimum initial horizontal velocity of the ball required to complete the full circle is approximately \(9.52 \, \text{m/s}\).