A charged particle of charge 'Q' is held fixed and another charged particle of mass 'm' and charge 'q' (of the same sign) is released from a distance 'r'. The impulse of the force exerted by the external agent on the fixed charge by the time distance between 'Q' and 'q' becomes `2 r` is

Options

A. `sqrt((Qq)/(4piepsilon_(0)mr))`
B. `sqrt((Qqm)/(4pi epsilon_(0)r))`
C. `sqrt((Qqm)/(pi epsilon_(0)r))`
D. `sqrt((Qqm)/(2pi epsilon_(0)r))`

Answer

To solve the problem step by step, we will analyze the situation using the principles of electrostatics and mechanics.

### Step 1: Understand the Problem
We have two charged particles:
- A fixed particle with charge \( Q \).
- A movable particle with charge \( q \) and mass \( m \) that is released from a distance \( r \) from the fixed charge.

Since both charges have the same sign, they will repel each other.

### Step 2: Identify the Forces Acting on the Movable Charge
When the charge \( q \) is released, it experiences a repulsive electrostatic force due to the charge \( Q \). The force \( F \) between the two charges can be calculated using Coulomb's Law:
\[ F = \frac{k \cdot |Q \cdot q|}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 3: Calculate the Initial and Final Energies
Initially, the potential energy \( U_i \) when the distance is \( r \) is given by:
\[ U_i = \frac{k \cdot Q \cdot q}{r} \]
The initial kinetic energy \( K_i \) is zero since the charge \( q \) is at rest.

When the distance becomes \( 2r \), the final potential energy \( U_f \) is:
\[ U_f = \frac{k \cdot Q \cdot q}{2r} \]
The final kinetic energy \( K_f \) can be expressed as:
\[ K_f = \frac{1}{2} m v^2 \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy:
\[ U_i + K_i = U_f + K_f \]
Substituting the values:
\[ \frac{k \cdot Q \cdot q}{r} + 0 = \frac{k \cdot Q \cdot q}{2r} + \frac{1}{2} m v^2 \]
Rearranging gives:
\[ \frac{k \cdot Q \cdot q}{r} - \frac{k \cdot Q \cdot q}{2r} = \frac{1}{2} m v^2 \]
\[ \frac{k \cdot Q \cdot q}{2r} = \frac{1}{2} m v^2 \]
Multiplying both sides by 2:
\[ \frac{k \cdot Q \cdot q}{r} = m v^2 \]
Thus, solving for \( v \):
\[ v = \sqrt{\frac{k \cdot Q \cdot q}{m r}} \]

### Step 5: Calculate the Change in Momentum
The initial momentum \( p_i \) is:
\[ p_i = 0 \quad (\text{since the charge is initially at rest}) \]
The final momentum \( p_f \) is:
\[ p_f = m v = m \sqrt{\frac{k \cdot Q \cdot q}{m r}} = \sqrt{m k Q q} \]
The change in momentum \( \Delta p \) is:
\[ \Delta p = p_f - p_i = \sqrt{m k Q q} - 0 = \sqrt{m k Q q} \]

### Step 6: Impulse Calculation
Impulse \( J \) is defined as the change in momentum:
\[ J = \Delta p = \sqrt{m k Q q} \]

### Conclusion
The impulse of the force exerted by the external agent on the fixed charge by the time the distance between \( Q \) and \( q \) becomes \( 2r \) is:
\[ J = \sqrt{m k Q q} \]