A particle of mass `m` carrying charge `q` is projected with velocity `v` from point `P` toward aninfinite line of charge from a distance `a` . Its speed reduces to zero momentarily at point `Q`, which is at a distance `a//2` from the line of charge. If another particle with mass `m` and charge `-q` is projected with the same velosity `v` from `P` toward the line of charge, what will be its speed at `Q` ?

Options

A. `v_1 = sqrt(2) v`
B. `v_1 = 2 v`
C. `v_1 = sqrtv`
D. `v_1 = sqrt(2)`

Explanation

To solve the problem, we will use the principle of conservation of energy. Here are the steps to find the speed of the second particle at point Q.

### Step 1: Understand the scenario
We have two particles:
1. Particle 1: Mass = m, Charge = q, Initial velocity = v, projected from point P at a distance a from an infinite line of charge.
2. Particle 2: Mass = m, Charge = -q, also projected from point P with the same initial velocity v.

Both particles are projected towards the line of charge, and we need to find the speed of the second particle at point Q, which is at a distance \( \frac{a}{2} \) from the line of charge.

### Step 2: Apply conservation of energy for Particle 1
For Particle 1, the change in potential energy (PE) and kinetic energy (KE) can be expressed as follows:
- Initial kinetic energy at point P:
\[ KE_{initial} = \frac{1}{2} mv^2 \]
- Potential energy at point P (distance a from the line of charge):
\[ PE_{initial} = k \frac{q \cdot Q}{a} \quad \text{(where Q is the linear charge density)} \]
- Potential energy at point Q (distance \( \frac{a}{2} \) from the line of charge):
\[ PE_{final} = k \frac{q \cdot Q}{\frac{a}{2}} = 2k \frac{q \cdot Q}{a} \]

Using conservation of energy:
\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]
Substituting the values:
\[ \frac{1}{2} mv^2 + k \frac{q \cdot Q}{a} = 0 + 2k \frac{q \cdot Q}{a} \]

### Step 3: Solve for Particle 1
Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = 2k \frac{q \cdot Q}{a} - k \frac{q \cdot Q}{a} \]
\[ \frac{1}{2} mv^2 = k \frac{q \cdot Q}{a} \]

### Step 4: Apply conservation of energy for Particle 2
For Particle 2, we will use the same approach:
- Initial kinetic energy at point P:
\[ KE_{initial} = \frac{1}{2} mv^2 \]
- Potential energy at point P:
\[ PE_{initial} = k \frac{-q \cdot Q}{a} \]
- Potential energy at point Q:
\[ PE_{final} = k \frac{-q \cdot Q}{\frac{a}{2}} = -2k \frac{q \cdot Q}{a} \]

Using conservation of energy:
\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]
Substituting the values:
\[ \frac{1}{2} mv^2 + k \frac{-q \cdot Q}{a} = KE_{final} + (-2k \frac{q \cdot Q}{a}) \]

### Step 5: Solve for Particle 2
Rearranging gives:
\[ \frac{1}{2} mv^2 + k \frac{q \cdot Q}{a} = KE_{final} \]
Substituting \( k \frac{q \cdot Q}{a} \) from Particle 1:
\[ KE_{final} = \frac{1}{2} mv^2 + k \frac{q \cdot Q}{a} \]
From Particle 1, we know:
\[ k \frac{q \cdot Q}{a} = \frac{1}{2} mv^2 \]
Thus:
\[ KE_{final} = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 \]

### Step 6: Find the final speed \( V_1 \)
Since \( KE_{final} = \frac{1}{2} m V_1^2 \):
\[ mv^2 = \frac{1}{2} m V_1^2 \]
\[ V_1^2 = 2v^2 \]
\[ V_1 = \sqrt{2}v \]

### Final Answer
The speed of the second particle at point Q is:
\[ V_1 = \sqrt{2}v \]