A thin wire carries a uniformly distributed charge Q = 3. 14nC along its length. The wire is bent in the form of a semi-circular arc of radius R = 1 m (without disturbing the charge distribution). Find the electric field at the center of curvature of the arc (in N/C).

Explanation

To find the electric field at the center of curvature of a semi-circular arc of radius \( R = 1 \, \text{m} \) carrying a uniformly distributed charge \( Q = 3.14 \, \text{nC} \), we can follow these steps:

### Step 1: Calculate the Linear Charge Density
The linear charge density \( \lambda \) is defined as the total charge per unit length. For a semi-circular arc, the length \( L \) of the arc is given by:
\[ L = \pi R \]
Substituting \( R = 1 \, \text{m} \):
\[ L = \pi \times 1 = \pi \, \text{m} \]
Now, we can find \( \lambda \):
\[ \lambda = \frac{Q}{L} = \frac{3.14 \times 10^{-9} \, \text{C}}{\pi \, \text{m}} = \frac{3.14 \times 10^{-9}}{3.14} = 10^{-9} \, \text{C/m} \]

### Step 2: Use the Electric Field Formula for a Semi-Circular Arc
The electric field \( E \) at the center of a semi-circular arc due to a uniformly charged wire is given by the formula:
\[ E = \frac{2k\lambda}{R} \]
where \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 3: Substitute the Values into the Formula
Substituting \( \lambda = 10^{-9} \, \text{C/m} \) and \( R = 1 \, \text{m} \):
\[ E = \frac{2 \times 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \times 10^{-9} \, \text{C/m}}{1 \, \text{m}} \]
This simplifies to:
\[ E = 2 \times 9 = 18 \, \text{N/C} \]

### Step 4: Conclusion
The electric field at the center of curvature of the semi-circular arc is:
\[ E = 18 \, \text{N/C} \]