A charged wire is bent in the form of a semicircular arc of radius a. If charge per unit length is `lambda` coulomb/metre, the electric field at the centre O is

Options

A. `(lambda)/(2pi a^(2)epsilon_(0))`
B. `(lambda)/(4pi^(2)epsilon_(0)a)`
C. `(lambda)/(2pi epsilon_(0)a)`
D. zero

Answer

To find the electric field at the center \( O \) of a semicircular arc of radius \( a \) with a uniform charge density \( \lambda \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
We have a semicircular arc with radius \( a \). The total charge is distributed uniformly along the arc.

2. **Consider a Small Element**:
Take a small charge element \( dq \) on the arc. The length of this small element is \( dl \), which can be expressed in terms of the angle \( d\theta \) as:
\[ dl = a \, d\theta \]
Therefore, the charge \( dq \) can be expressed as:
\[ dq = \lambda \, dl = \lambda \, a \, d\theta \]

3. **Electric Field Contribution**:
The electric field \( dE \) due to the charge element \( dq \) at the center \( O \) can be calculated using Coulomb's law:
\[ dE = \frac{k \, dq}{a^2} = \frac{k \, \lambda \, a \, d\theta}{a^2} = \frac{k \, \lambda \, d\theta}{a} \]
where \( k = \frac{1}{4\pi \epsilon_0} \).

4. **Components of Electric Field**:
The electric field \( dE \) can be resolved into two components:
- The radial component \( dE \cos \theta \) (which will cancel out due to symmetry).
- The vertical component \( dE \sin \theta \) (which will add up).

5. **Calculate the Vertical Component**:
The vertical component of the electric field \( dE_y \) is given by:
\[ dE_y = dE \sin \theta = \frac{k \lambda}{a} \sin \theta \, d\theta \]

6. **Integrate Over the Semicircular Arc**:
We need to integrate \( dE_y \) from \( \theta = 0 \) to \( \theta = \pi \):
\[ E_y = \int_0^{\pi} dE_y = \int_0^{\pi} \frac{k \lambda}{a} \sin \theta \, d\theta \]
The integral of \( \sin \theta \) is:
\[ \int \sin \theta \, d\theta = -\cos \theta \]
Evaluating from \( 0 \) to \( \pi \):
\[ E_y = \frac{k \lambda}{a} \left[-\cos \theta \right]_0^{\pi} = \frac{k \lambda}{a} \left[-(-1 - 1)\right] = \frac{k \lambda}{a} \cdot 2 = \frac{2 k \lambda}{a} \]

7. **Substituting for \( k \)**:
Substitute \( k = \frac{1}{4\pi \epsilon_0} \):
\[ E_y = \frac{2 \cdot \frac{1}{4\pi \epsilon_0} \lambda}{a} = \frac{\lambda}{2\pi \epsilon_0 a} \]

### Final Result:
The electric field at the center \( O \) of the semicircular arc is:
\[ E = \frac{\lambda}{2\pi \epsilon_0 a} \]