To find the electric field at the center of a uniformly charged semicircular arc of radius \( a \) with a charge per unit length \( \lambda \), we can follow these steps:
### Step 1: Understand the Charge Distribution
The semicircular arc has a uniform charge distribution. The total charge \( Q \) on the arc can be expressed in terms of the charge per unit length \( \lambda \) and the length of the arc. The length \( L \) of the semicircular arc is given by:
\[
L = \pi a
\]
Thus, the total charge \( Q \) is:
\[
Q = \lambda L = \lambda \pi a
\]
### Step 2: Consider an Element of Charge
To find the electric field at the center, we consider a small element of charge \( dQ \) on the arc. The length of this small element \( dL \) can be expressed in terms of an angle \( d\theta \):
\[
dL = a \, d\theta
\]
Therefore, the charge \( dQ \) corresponding to this element is:
\[
dQ = \lambda \, dL = \lambda a \, d\theta
\]
### Step 3: Calculate the Electric Field Contribution
The electric field \( dE \) due to the charge \( dQ \) at the center of the semicircle (point O) can be expressed using Coulomb's law:
\[
dE = \frac{k \, dQ}{a^2} = \frac{k \, \lambda a \, d\theta}{a^2} = \frac{k \lambda}{a} \, d\theta
\]
where \( k \) is Coulomb's constant.
### Step 4: Resolve the Electric Field Components
The electric field \( dE \) has both x and y components. The x-component \( dE_x \) and the y-component \( dE_y \) can be expressed as:
\[
dE_x = dE \cos \theta = \frac{k \lambda}{a} \, d\theta \cos \theta
\]
\[
dE_y = dE \sin \theta = \frac{k \lambda}{a} \, d\theta \sin \theta
\]
### Step 5: Integrate the Electric Field Components
Since the semicircular arc is symmetric, the y-components \( dE_y \) from opposite sides will cancel each other out. Thus, we only need to integrate the x-component \( dE_x \):
\[
E_x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dE_x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{k \lambda}{a} \cos \theta \, d\theta
\]
This integral can be simplified as:
\[
E_x = \frac{k \lambda}{a} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \, d\theta
\]
### Step 6: Evaluate the Integral
The integral of \( \cos \theta \) from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) is:
\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \, d\theta = [\sin \theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) = 1 - (-1) = 2
\]
Thus,
\[
E_x = \frac{k \lambda}{a} \cdot 2 = \frac{2k \lambda}{a}
\]
### Step 7: Substitute Coulomb's Constant
Substituting \( k = \frac{1}{4 \pi \epsilon_0} \):
\[
E_x = \frac{2 \cdot \frac{1}{4 \pi \epsilon_0} \lambda}{a} = \frac{\lambda}{2 \pi \epsilon_0 a}
\]
### Final Result
The electric field at the center of the semicircular arc is:
\[
E = \frac{\lambda}{2 \pi \epsilon_0 a}
\]
