A charge particle 'q' is shot towards another charged particle 'Q' which is fixed, with a speed 'v'. It approaches 'Q' upto a closest distance r and then returns. If q were given a speed of '2v' the closest distances of approach would be

Options

A. (a) `r/2`
B. (b) `2/r`
C. (c) `r`
D. (d) `r/4`

Explanation

To solve the problem, we need to apply the principle of conservation of energy. The total mechanical energy of the system (kinetic energy + potential energy) remains constant as the charged particle approaches the fixed charge.

### Step-by-Step Solution:

1. **Initial Conditions**:
- A charge particle \( q \) is shot towards a fixed charge \( Q \) with an initial speed \( v \).
- The closest distance of approach is \( r \).

2. **Energy at Closest Distance**:
- The initial kinetic energy \( KE_1 \) when the particle is shot is given by:
\[ KE_1 = \frac{1}{2} mv^2 \]
- The potential energy \( PE \) at the closest distance \( r \) is given by:
\[ PE = \frac{kQq}{r} \]
- By conservation of energy, we have:
\[ KE_1 = PE \]
- Therefore, we can write:
\[ \frac{1}{2} mv^2 = \frac{kQq}{r} \quad \text{(1)} \]

3. **New Conditions with Speed \( 2v \)**:
- Now, if the charge particle \( q \) is given a speed of \( 2v \), the new kinetic energy \( KE_2 \) becomes:
\[ KE_2 = \frac{1}{2} m(2v)^2 = \frac{1}{2} m \cdot 4v^2 = 2mv^2 \]
- Let the new closest distance of approach be \( r' \). The potential energy at this distance is:
\[ PE' = \frac{kQq}{r'} \]
- By conservation of energy, we have:
\[ KE_2 = PE' \]
- Therefore, we can write:
\[ 2mv^2 = \frac{kQq}{r'} \quad \text{(2)} \]

4. **Relating the Two Equations**:
- From equation (1), we know:
\[ \frac{kQq}{r} = \frac{1}{2} mv^2 \]
- Substituting this into equation (2):
\[ 2mv^2 = 4 \cdot \frac{kQq}{r} \]
- Rearranging gives:
\[ \frac{kQq}{r'} = 4 \cdot \frac{kQq}{r} \]
- This implies:
\[ r' = \frac{r}{4} \]

5. **Conclusion**:
- The closest distance of approach when the charge particle \( q \) is given a speed of \( 2v \) is:
\[ r' = \frac{r}{4} \]

### Final Answer:
The closest distance of approach when the speed is \( 2v \) is \( \frac{r}{4} \).

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