A particle A of charge ` 1 mu C` is held fixed at a point P in free space . Another particle B of same charge and mass `4 mu g` is kept at a distance of1mm from P .If B is released then its velocity at a distance of 9 mm from P is (Take `(1)/( 4pi epsilon_(0)) =9 xx 10^(9)Nm^(2) C^(-2))`

Options

A. `2.0xx10^3m//s`
B. `1.5xx10^2m//s`
C. `3.0xx10^4m//s`
D. `1.0 m//s`

Explanation

To solve the problem step by step, we will use the principle of conservation of mechanical energy. The total mechanical energy (potential energy + kinetic energy) remains constant since there are no external forces acting on the system.

### Step 1: Understand the Initial Conditions
- Particle A has a charge \( Q_A = 1 \mu C = 1 \times 10^{-6} C \) and is fixed at point P.
- Particle B has the same charge \( Q_B = 1 \mu C = 1 \times 10^{-6} C \) and mass \( m = 4 \mu g = 4 \times 10^{-9} kg \).
- Particle B is initially at a distance \( r_i = 1 mm = 1 \times 10^{-3} m \) from P and is released from rest.

### Step 2: Write the Conservation of Energy Equation
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