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. `0.33xx10^(11)N//C`
B. `3xx10^(11)NC^(-1)`
C. `0.66xx10^(11)NC^(-1)`
D. `1.32xx10^(11)NC^(-1)`

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Linear charge density, \( \lambda = \frac{1}{3} \, \text{C/m} \)
- Distance from the wire, \( r = 18 \, \text{cm} = 0.18 \, \text{m} \)

2. **Use the Formula for Electric Field:**
The electric field \( E \) due to an infinitely long line charge is given by the formula:
\[ E = \frac{2k\lambda}{r} \]
where \( k \) is Coulomb's constant, approximately \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

3. **Substitute the Values into the Formula:**
\[ E = \frac{2 \times (9 \times 10^9) \times \left(\frac{1}{3}\right)}{0.18} \]

4. **Calculate the Electric Field:**
- First, calculate \( 2 \times 9 \times 10^9 \):
\[ 2 \times 9 \times 10^9 = 18 \times 10^9 \]
- Now substitute this into the equation:
\[ E = \frac{18 \times 10^9 \times \frac{1}{3}}{0.18} \]
- Simplify \( \frac{18}{3} = 6 \):
\[ E = \frac{6 \times 10^9}{0.18} \]
- Now calculate \( \frac{6 \times 10^9}{0.18} \):
\[ E = 33.33 \times 10^9 \, \text{N/C} \]

5. **Final Result:**
The magnitude of the electric field intensity at a point 18 cm away from the wire is:
\[ E \approx 3.33 \times 10^{10} \, \text{N/C} \]