A very long, straight, thin wire carries `-3.60 nCm^(-1)` of fixed negative charge. The wire is to be surrounded by a uniform cylinder of positive charge, radius 1.50 cm, coaxial with the wire. The volume charge density `rho` of the cylinder is to be selected so that the net electric field outside the cylinder is zero. Calculate the required positive charge density `rho` (in `muCm^(-3)`).

Options

A. 6
B. 7
C. 5
D. 3

Explanation

To solve the problem, we need to find the volume charge density \( \rho \) of the positive charge cylinder that surrounds a long straight wire with a negative linear charge density \( \lambda \). The goal is to ensure that the net electric field outside the cylinder is zero.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Linear charge density of the wire: \( \lambda = -3.60 \, \text{nC/m} = -3.60 \times 10^{-9} \, \text{C/m} \)
- Radius of the cylinder: \( r = 1.50 \, \text{cm} = 1.50 \times 10^{-2} \, \text{m} \)

2. **Electric Field Due to the Wire:**
The electric field \( E \) at a distance \( r \) from a long straight wire with linear charge density \( \lambda \) is given by:
\[ E_{\text{wire}} = \frac{\lambda}{2 \pi \epsilon_0 r} \]
where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

3. **Electric Field Due to the Cylinder:**
The electric field outside a uniformly charged cylinder with linear charge density \( \lambda' \) is given by:
\[ E_{\text{cylinder}} = \frac{\lambda'}{2 \pi \epsilon_0 r} \]

4. **Condition for Zero Net Electric Field:**
For the net electric field outside the cylinder to be zero, the electric fields due to the wire and the cylinder must be equal in magnitude and opposite in direction:
\[ E_{\text{wire}} + E_{\text{cylinder}} = 0 \]
This implies:
\[ \frac{\lambda}{2 \pi \epsilon_0 r} + \frac{\lambda'}{2 \pi \epsilon_0 r} = 0 \]
Simplifying gives:
\[ \lambda + \lambda' = 0 \quad \Rightarrow \quad \lambda' = -\lambda \]

5. **Relating Linear Charge Density to Volume Charge Density:**
The linear charge density \( \lambda' \) of the cylinder can be expressed in terms of the volume charge density \( \rho \) as follows:
\[ \lambda' = \rho \cdot V \]
where \( V \) is the volume of the cylinder. The volume \( V \) of a cylinder can be calculated as:
\[ V = \pi r^2 L \]
where \( L \) is the length of the cylinder. Thus,
\[ \lambda' = \rho \cdot \pi r^2 L \]

6. **Substituting for \( \lambda' \):**
From the previous step, we have:
\[ -\lambda = \rho \cdot \pi r^2 \]
Rearranging gives:
\[ \rho = -\frac{\lambda}{\pi r^2} \]

7. **Calculating \( \rho \):**
Substitute the values for \( \lambda \) and \( r \):
\[ \rho = -\frac{-3.60 \times 10^{-9}}{\pi (1.50 \times 10^{-2})^2} \]
Calculate \( \pi (1.50 \times 10^{-2})^2 \):
\[ \pi (1.50 \times 10^{-2})^2 \approx 3.14 \times (2.25 \times 10^{-4}) \approx 7.068 \times 10^{-4} \]
Now substituting this back:
\[ \rho = \frac{3.60 \times 10^{-9}}{7.068 \times 10^{-4}} \approx 5.09 \times 10^{-6} \, \text{C/m}^3 \]

8. **Convert to Microcoulombs per Cubic Meter:**
\[ \rho \approx 5.09 \, \mu\text{C/m}^3 \]

### Final Answer:
The required positive charge density \( \rho \) is approximately \( 5.09 \, \mu\text{C/m}^3 \).