Charge `q` is uniformly distributed over a thin half ring of radius `R`. The electric field at the centre of the ring is

Options

A. `(q)/(2 pi^(2) epsi_(0)R^(2))`
B. `(q)/(4 pi^(2) epsi_(0)R^(2))`
C. `(q)/(4 pi rpsi_(0)R^(2))`
D. `(q)/(2 pi epsi_(0)R^(2))`

Answer

To find the electric field at the center of a thin half ring of radius \( R \) with a uniformly distributed charge \( Q \), we can follow these steps:

### Step 1: Define the Linear Charge Density
The charge \( Q \) is uniformly distributed over the half ring. The length of the half ring (semi-circumference) is given by:
\[ L = \pi R \]
The linear charge density \( \lambda \) is defined as:
\[ \lambda = \frac{Q}{L} = \frac{Q}{\pi R} \]

### Step 2: Consider a Small Charge Element
Let’s consider a small length element \( ds \) on the half ring. The charge \( dQ \) on this small element can be expressed as:
\[ dQ = \lambda \, ds \]
Since \( ds = R \, d\theta \) (where \( d\theta \) is the angle subtended at the center by the small element), we have:
\[ dQ = \lambda \, R \, d\theta = \frac{Q}{\pi R} \cdot R \, d\theta = \frac{Q}{\pi} \, d\theta \]

### Step 3: Calculate the Electric Field Contribution
The electric field \( dE \) due to the small charge \( dQ \) at the center of the ring can be expressed using Coulomb's law:
\[ dE = \frac{dQ}{4 \pi \epsilon_0 R^2} \]
Substituting \( dQ \):
\[ dE = \frac{1}{4 \pi \epsilon_0 R^2} \cdot \frac{Q}{\pi} \, d\theta = \frac{Q}{4 \pi^2 \epsilon_0 R^2} \, d\theta \]

### Step 4: Resolve Electric Field into Components
The electric field \( dE \) can be resolved into two components: \( dE_x \) and \( dE_y \). Due to symmetry, the horizontal components (x-components) from opposite sides will cancel each other out, leaving only the vertical components (y-components) contributing to the net electric field:
\[ dE_y = dE \sin \theta \]
Thus, the total vertical component from both sides is:
\[ dE_{net} = 2 dE_y = 2 dE \sin \theta \]

### Step 5: Substitute and Integrate
Substituting for \( dE \):
\[ dE_{net} = 2 \cdot \frac{Q}{4 \pi^2 \epsilon_0 R^2} \cdot \sin \theta \, d\theta \]
Now, we need to integrate this from \( \theta = 0 \) to \( \theta = \pi \):
\[ E = \int_0^{\pi} 2 \cdot \frac{Q}{4 \pi^2 \epsilon_0 R^2} \cdot \sin \theta \, d\theta \]
Calculating the integral:
\[ E = \frac{Q}{2 \pi^2 \epsilon_0 R^2} \cdot \left[-\cos \theta \right]_0^{\pi} = \frac{Q}{2 \pi^2 \epsilon_0 R^2} \cdot \left[-\cos(\pi) + \cos(0)\right] \]
\[ = \frac{Q}{2 \pi^2 \epsilon_0 R^2} \cdot [1 + 1] = \frac{Q}{\pi^2 \epsilon_0 R^2} \]

### Final Result
Thus, the electric field at the center of the half ring is:
\[ E = \frac{Q}{2 \pi \epsilon_0 R^2} \]