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 a 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 the charge per unit length. Thus, for a small arc length \( d\theta \) at radius \( R \):

\[ dq = \lambda \cdot dL = \lambda \cdot R \cdot d\theta \]

### Step 2: Set Up the Electric Field Contribution
The electric field \( dE \) due to the charge element \( dq \) at the center of the semicircular ring can be expressed using Coulomb's law:

\[ dE = \frac{k \cdot dq}{R^2} \]

Where \( k \) is Coulomb's constant, \( k = \frac{1}{4\pi\epsilon_0} \).

Substituting for \( dq \):

\[ dE = \frac{k \cdot \lambda \cdot R \cdot d\theta}{R^2} = \frac{k \cdot \lambda}{R} \cdot d\theta \]

### Step 3: Resolve the Electric Field Components
The electric field \( dE \) has both horizontal (x-direction) and vertical (y-direction) components. Due to symmetry, the horizontal components from opposite sides will cancel out, and we only need to consider the vertical components.

The vertical component \( dE_y \) is given by:

\[ dE_y = dE \cdot \sin\theta \]

Where \( \theta \) is the angle corresponding to the position of the charge element on the semicircle.

### Step 4: Substitute for \( dE \)
Substituting for \( dE \):

\[ dE_y = \frac{k \cdot \lambda}{R} \cdot d\theta \cdot \sin\theta \]

### Step 5: Integrate Over the Semicircle
To find the total electric field \( E_y \) at the center, integrate \( dE_y \) from \( \theta = 0 \) to \( \theta = \pi \):

\[ E_y = \int_0^{\pi} dE_y = \int_0^{\pi} \frac{k \cdot \lambda}{R} \cdot \sin\theta \, d\theta \]

### Step 6: Evaluate the Integral
The integral of \( \sin\theta \) is:

\[ \int \sin\theta \, d\theta = -\cos\theta \]

Thus, evaluating the definite integral:

\[ E_y = \frac{k \cdot \lambda}{R} \left[-\cos\theta\right]_0^{\pi} = \frac{k \cdot \lambda}{R} \left[-(-1) - (-1)\right] = \frac{k \cdot \lambda}{R} \cdot 2 \]

### Step 7: Final Expression for Electric Field
Substituting \( k = \frac{1}{4\pi\epsilon_0} \):

\[ E_y = \frac{2k \cdot \lambda}{R} = \frac{2 \cdot \frac{1}{4\pi\epsilon_0} \cdot \lambda}{R} = \frac{\lambda}{2\pi\epsilon_0 R} \]

### Conclusion
The electric field at the center of the uniformly charged semicircular ring is:

\[ E = \frac{\lambda}{2\pi\epsilon_0 R} \]