An infinitely long thin straight wire has uniform linear charge density of `1//3 "coul" m^(-1)` .Then the magnitude of the electric intensity at a point `18cm` away is

Options

A. `0.33 xx 10^(11) NC^(-1)`
B. `3 xx 10^(11) NC^(-1)`
C. `0.66 xx 10^(11) NC^(-1)`
D. `1.32 xx 10^(11) NC^(-1)`

Answer

To find the electric field intensity (E) at a point 18 cm away from an infinitely long thin straight wire with a uniform linear charge density (λ) of \( \frac{1}{3} \, \text{C/m} \), we can use the formula for the electric field due to an infinitely long charged wire:

\[ E = \frac{\lambda}{2 \pi \epsilon_0 R} \]

where:
- \( E \) is the electric field intensity,
- \( \lambda \) is the linear charge density,
- \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)),
- \( R \) is the distance from the wire.

### Step 1: Convert the distance from centimeters to meters
Given distance \( R = 18 \, \text{cm} \):
\[ R = 18 \, \text{cm} = 18 \times 10^{-2} \, \text{m} = 0.18 \, \text{m} \]

### Step 2: Substitute the values into the formula
Given:
- \( \lambda = \frac{1}{3} \, \text{C/m} \)
- \( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

Now, substitute these values into the electric field formula:
\[ E = \frac{\frac{1}{3}}{2 \pi (8.85 \times 10^{-12}) (0.18)} \]

### Step 3: Calculate the denominator
First, calculate \( 2 \pi \epsilon_0 R \):
\[ 2 \pi \epsilon_0 R = 2 \pi (8.85 \times 10^{-12}) (0.18) \]
Calculating this gives:
\[ 2 \pi (8.85 \times 10^{-12}) (0.18) \approx 1.00 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \]

### Step 4: Calculate the electric field intensity
Now substitute back into the formula:
\[ E = \frac{\frac{1}{3}}{1.00 \times 10^{-12}} \approx \frac{0.333}{1.00 \times 10^{-12}} \approx 3.33 \times 10^{11} \, \text{N/m}^2 \]

### Final Answer
The magnitude of the electric field intensity at a point 18 cm away from the wire is approximately:
\[ E \approx 3.33 \times 10^{11} \, \text{N/m}^2 \]