A long straight wire carries a charge with linear density `lamda`. A particle of mass m and a charge q is released at a distance r from the the wire. The speed of the particle as it crosses a point distance 2r is

Options

A. `sqrt((q lamda ln r)/(pi m in_(0)))`
B. `sqrt((q lamda ln2)/(pi m in_(0)))`
C. `sqrt((q lamda ln2)/(2 pi m in_(0)))`
D. `sqrt((q lamda ln2)/(pim in_(0)))`

Explanation

To solve the problem, we need to determine the speed of a charged particle as it moves from a distance \( r \) to a distance \( 2r \) from a long straight wire that has a linear charge density \( \lambda \). We will use the concepts of electric field, work done, and the work-energy theorem.

### Step-by-Step Solution:

1. **Electric Field due to the Wire**:
The electric field \( E \) at a distance \( x \) from a long straight wire with linear charge density \( \lambda \) is given by:
\[ E = \frac{\lambda}{2 \pi \epsilon_0 x} \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Force on the Particle**:
The force \( F \) acting on the particle of charge \( q \) in the electric field is:
\[ F = qE = q \cdot \frac{\lambda}{2 \pi \epsilon_0 x} \]

3. **Work Done by the Electric Force**:
The work done \( W \) by the electric force when the particle moves from distance \( r \) to \( 2r \) can be calculated by integrating the force over the distance:
\[ W = \int_{r}^{2r} F \, dx = \int_{r}^{2r} q \cdot \frac{\lambda}{2 \pi \epsilon_0 x} \, dx \]
This simplifies to:
\[ W = \frac{q \lambda}{2 \pi \epsilon_0} \int_{r}^{2r} \frac{1}{x} \, dx \]
The integral of \( \frac{1}{x} \) is \( \ln x \), so:
\[ W = \frac{q \lambda}{2 \pi \epsilon_0} \left[ \ln(2r) - \ln(r) \right] = \frac{q \lambda}{2 \pi \epsilon_0} \ln\left(\frac{2r}{r}\right) = \frac{q \lambda}{2 \pi \epsilon_0} \ln(2) \]

4. **Using the Work-Energy Theorem**:
According to the work-energy theorem, the work done on the particle is equal to the change in kinetic energy. Since the particle starts from rest, the initial kinetic energy is zero:
\[ W = \Delta KE = KE_f - KE_i = \frac{1}{2} mv^2 - 0 \]
Therefore:
\[ \frac{q \lambda}{2 \pi \epsilon_0} \ln(2) = \frac{1}{2} mv^2 \]

5. **Solving for the Speed \( v \)**:
Rearranging the equation to solve for \( v \):
\[ mv^2 = \frac{2q \lambda}{2 \pi \epsilon_0} \ln(2) \]
\[ v^2 = \frac{q \lambda \ln(2)}{m \pi \epsilon_0} \]
\[ v = \sqrt{\frac{q \lambda \ln(2)}{m \pi \epsilon_0}} \]

### Final Answer:
The speed of the particle as it crosses the point at a distance \( 2r \) from the wire is:
\[ v = \sqrt{\frac{q \lambda \ln(2)}{m \pi \epsilon_0}} \]