A small ball of mass `2 xx 10^-3 kg`, having a charge `1 mu C`, is suspended by a string of length `0.8 m`. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity that should be imparted to the lower ball so that it can make a complete revolution.

Explanation

To solve the problem of determining the minimum horizontal velocity that should be imparted to the lower ball so that it can make a complete revolution, we can follow these steps:

### Step 1: Understand the Forces Acting on the Ball
When the lower ball is at the top of its circular path, the forces acting on it are:
- The gravitational force (downward) = \( mg \)
- The electrostatic repulsive force due to the upper ball (also downward) = \( \frac{k q^2}{L^2} \)

Where:
- \( m = 2 \times 10^{-3} \, \text{kg} \) (mass of the ball)
- \( q = 1 \times 10^{-6} \, \text{C} \) (charge of the ball)
- \( L = 0.8 \, \text{m} \) (length of the string)
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant)

### Step 2: Apply the Condition for Circular Motion
For the ball to just complete the circular motion, the centripetal force required must equal the net force acting towards the center. At the top of the circular path, the tension in the string can be considered zero for minimum velocity. Therefore, we have:

\[ \frac{mv^2}{L} = mg + \frac{kq^2}{L^2} \]

Where \( v \) is the velocity at the top of the circular path.

### Step 3: Rearrange the Equation
Rearranging the equation gives:

\[ v^2 = gL + \frac{kq^2}{m} \]

### Step 4: Calculate the Velocity at the Top
Now substituting the values:

1. Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)
2. \( m = 2 \times 10^{-3} \, \text{kg} \)
3. \( q = 1 \times 10^{-6} \, \text{C} \)
4. \( L = 0.8 \, \text{m} \)

Calculating the electrostatic force:

\[ \frac{kq^2}{L^2} = \frac{(9 \times 10^9)(1 \times 10^{-6})^2}{(0.8)^2} = \frac{9 \times 10^9 \times 10^{-12}}{0.64} = \frac{9 \times 10^{-3}}{0.64} \approx 0.0140625 \, \text{N} \]

Now substituting into the equation for \( v^2 \):

\[ v^2 = (10)(0.8) + \frac{9 \times 10^{-3}}{2 \times 10^{-3}} = 8 + 4.5 = 12.5 \]

### Step 5: Calculate the Minimum Velocity
Now, taking the square root to find \( v \):

\[ v = \sqrt{12.5} \approx 3.54 \, \text{m/s} \]

### Step 6: Consider Energy Conservation
To find the velocity \( v_b \) at the bottom, we can use conservation of energy. The potential energy at the top will equal the kinetic energy at the bottom plus the potential energy at the bottom.

\[ \frac{1}{2}mv_b^2 = \frac{1}{2}mv^2 + mg(2L) \]

Substituting \( v^2 \) from earlier:

\[ v_b^2 = v^2 + 4gL \]

Substituting values:

\[ v_b^2 = 12.5 + 4(10)(0.8) = 12.5 + 32 = 44.5 \]

Taking the square root gives:

\[ v_b = \sqrt{44.5} \approx 6.67 \, \text{m/s} \]

### Final Answer
The minimum horizontal velocity that should be imparted to the lower ball so that it can make a complete revolution is approximately **6.67 m/s**.

---