Find the electric field at the centre of a uniformly charged semicircular ring of radius R. Linear charge density is `lamda`

Explanation

To find the electric field at the center of a uniformly charged semicircular ring of radius \( R \) with linear charge density \( \lambda \), we can follow these steps:

### Step 1: Define the Charge Element
Consider a small charge element \( dq \) on the semicircular ring. The linear charge density \( \lambda \) is defined as charge per unit length. Therefore, for an infinitesimal arc length \( dl \), we have:
\[ dq = \lambda \, dl \]
For a semicircular ring, the arc length \( dl \) can be expressed in terms of the angle \( \theta \) as:
\[ dl = R \, d\theta \]
Thus, we can write:
\[ dq = \lambda \, R \, d\theta \]

### Step 2: Determine the Electric Field Contribution
The electric field \( dE \) due to the charge element \( dq \) at the center of the semicircle can be expressed using Coulomb's law:
\[ dE = \frac{k \, dq}{R^2} \]
Substituting \( dq \):
\[ dE = \frac{k \, \lambda \, R \, d\theta}{R^2} = \frac{k \, \lambda \, d\theta}{R} \]
where \( k \) is Coulomb's constant.

### Step 3: Resolve the Electric Field Components
The electric field \( dE \) has both x and y components. Due to symmetry, the x-components from opposite sides will cancel out. Therefore, we only need to consider the y-component:
\[ dE_y = dE \cos(\theta) = \frac{k \, \lambda \, d\theta}{R} \cos(\theta) \]

### Step 4: Integrate to Find the Total Electric Field
To find the total electric field \( E_y \) at the center, we integrate \( dE_y \) from \( \theta = 0 \) to \( \theta = \pi \):
\[ E_y = \int_0^{\pi} dE_y = \int_0^{\pi} \frac{k \, \lambda}{R} \cos(\theta) \, d\theta \]
Calculating the integral:
\[ E_y = \frac{k \, \lambda}{R} \int_0^{\pi} \cos(\theta) \, d\theta \]
The integral of \( \cos(\theta) \) from \( 0 \) to \( \pi \) is:
\[ \int_0^{\pi} \cos(\theta) \, d\theta = [\sin(\theta)]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \]
However, we need to consider the contributions from both halves of the semicircle, which gives us:
\[ E_y = 2 \cdot \frac{k \, \lambda}{R} \int_0^{\frac{\pi}{2}} \cos(\theta) \, d\theta \]
Calculating this integral:
\[ \int_0^{\frac{\pi}{2}} \cos(\theta) \, d\theta = [\sin(\theta)]_0^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]
Thus:
\[ E_y = 2 \cdot \frac{k \, \lambda}{R} \cdot 1 = \frac{2k \, \lambda}{R} \]

### Final Result
The electric field at the center of the uniformly charged semicircular ring is:
\[ E = \frac{2k \, \lambda}{R} \]