Two identical particles of mass m carry a charge `Q`, each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v. The closest distance of approach be .

Options

A. `(1)/( 4 pi epsi_(0)) (Q^(2))/(m v)`
B. `(1)/(4 pi epsi_(0)) (4Q^(2))/(m v^(2))`
C. `(1)/(4 pi epsi_(0)) (2Q^(2))/(m v^(2))`
D. `(1)/(4 pi epsi_(0))(3Q^(2))/(m v^(2))`

Explanation

To solve the problem, we will use the principles of conservation of momentum and conservation of energy. Let's break down the solution step by step.

### Step 1: Understand the initial conditions
We have two identical particles, each with mass \( m \) and charge \( Q \). One particle is at rest, while the other is projected towards it with speed \( v \).

### Step 2: Apply conservation of momentum
At the closest distance of approach, both particles will be moving due to the repulsive electrostatic force between them. Let the speed of each particle at this point be \( u \). According to the conservation of momentum:

\[ mv + 0 = mu + mu \]

This simplifies to:

\[ mv = 2mu \]

From this, we can solve for \( u \):

\[ u = \frac{v}{2} \]

### Step 3: Apply conservation of energy
At the closest distance of approach, the total mechanical energy is conserved. The initial kinetic energy of the moving particle is:

\[ \text{Initial KE} = \frac{1}{2} mv^2 \]

At the closest distance, the kinetic energy of both particles is:

\[ \text{Final KE} = \frac{1}{2} m u^2 + \frac{1}{2} m u^2 = m \left(\frac{1}{2} u^2\right) = m \left(\frac{1}{2} \left(\frac{v}{2}\right)^2\right) = \frac{mv^2}{8} \]

Additionally, at the closest distance \( d \), the potential energy due to the electrostatic force is given by:

\[ \text{Potential Energy} = \frac{1}{4\pi \epsilon_0} \frac{Q^2}{d} \]

### Step 4: Set up the energy conservation equation
According to the conservation of energy:

\[ \text{Initial KE} = \text{Final KE} + \text{Potential Energy} \]

Substituting the values we have:

\[ \frac{1}{2} mv^2 = \frac{mv^2}{8} + \frac{1}{4\pi \epsilon_0} \frac{Q^2}{d} \]

### Step 5: Simplify the equation
Rearranging the equation gives:

\[ \frac{1}{2} mv^2 - \frac{mv^2}{8} = \frac{1}{4\pi \epsilon_0} \frac{Q^2}{d} \]

Calculating the left side:

\[ \frac{4}{8} mv^2 - \frac{1}{8} mv^2 = \frac{3}{8} mv^2 \]

Thus, we have:

\[ \frac{3}{8} mv^2 = \frac{1}{4\pi \epsilon_0} \frac{Q^2}{d} \]

### Step 6: Solve for \( d \)
Rearranging for \( d \):

\[ d = \frac{1}{4\pi \epsilon_0} \frac{Q^2}{\frac{3}{8} mv^2} \]

This simplifies to:

\[ d = \frac{8Q^2}{12\pi \epsilon_0 mv^2} = \frac{2Q^2}{3\pi \epsilon_0 mv^2} \]

### Final Answer
The closest distance of approach \( d \) is:

\[ d = \frac{2Q^2}{3\pi \epsilon_0 mv^2} \]