A charge moves with a speed `v` in a circular path of radius `r` around a long uniformly charged conductor.

Options

A. `v prop r`
B. `v prop (1)/(r )`
C. `v prop (1)/(sqrt(r ))`
D. `v prop r^(0)`

Explanation

To solve the problem of how the velocity \( v \) of a charge moving in a circular path around a long uniformly charged conductor depends on the radius \( r \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Charge:**
The charge \( q \) moving in a circular path experiences a centripetal force that is provided by the electric force due to the uniformly charged conductor.

2. **Centripetal Force Equation:**
The centripetal force required to keep the charge moving in a circular path is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the charge and \( v \) is its speed.

3. **Electric Field Due to the Charged Conductor:**
The electric field \( E \) at a distance \( r \) from a long uniformly charged conductor with linear charge density \( \lambda \) is given by:
\[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \]
where \( \epsilon_0 \) is the permittivity of free space.

4. **Electric Force on the Charge:**
The electric force \( F_e \) acting on the charge \( q \) due to the electric field is:
\[ F_e = qE = q \left(\frac{\lambda}{2 \pi \epsilon_0 r}\right) \]

5. **Set the Forces Equal:**
For circular motion, the centripetal force is provided by the electric force:
\[ \frac{mv^2}{r} = q \left(\frac{\lambda}{2 \pi \epsilon_0 r}\right) \]

6. **Simplify the Equation:**
Multiplying both sides by \( r \) gives:
\[ mv^2 = q \left(\frac{\lambda}{2 \pi \epsilon_0}\right) \]

7. **Solve for \( v^2 \):**
Rearranging the equation to isolate \( v^2 \):
\[ v^2 = \frac{q \lambda}{2 \pi \epsilon_0 m} \]

8. **Conclusion about the Dependence of \( v \) on \( r \):**
Notice that in the equation for \( v^2 \), there is no \( r \) term present. This indicates that the speed \( v \) does not depend on the radius \( r \) of the circular path.

9. **Final Expression for \( v \):**
Taking the square root gives:
\[ v = \sqrt{\frac{q \lambda}{2 \pi \epsilon_0 m}} \]
Thus, \( v \) is independent of \( r \).

### Final Answer:
The speed \( v \) of the charge moving in a circular path around a long uniformly charged conductor does not depend on the radius \( r \).