Charges `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)/(2pi^(2)epsilon_(0)R^(2))`
B. `(q)/(4pi^(2)epsilon_(0)R^(2))`
C. `(q)/(4piepsilon_(0)R^(2))`
D. `(q)/(2piepsilon_(0)R^(2))`

Explanation

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 Charge Distribution
The total charge \( q \) is uniformly distributed over the half ring. The length of the half ring is \( \pi R \).

### Step 2: Determine Charge per Unit Length
The charge per unit length \( \lambda \) on the half ring can be calculated as:
\[ \lambda = \frac{q}{\pi R} \]

### Step 3: Consider a Small Element of Charge
Take a small element of the half ring, \( dL \), at an angle \( \theta \) from the vertical. The length of this small element can be expressed as:
\[ dL = R \, d\theta \]
The charge \( dQ \) on this small element is:
\[ dQ = \lambda \, dL = \frac{q}{\pi R} \cdot R \, d\theta = \frac{q}{\pi} \, d\theta \]

### Step 4: Calculate the Electric Field Due to \( dQ \)
The electric field \( dE \) at the center of the ring due to the charge \( dQ \) is given by Coulomb's law:
\[ dE = k \frac{dQ}{R^2} = k \frac{\frac{q}{\pi} \, d\theta}{R^2} = \frac{kq}{\pi R^2} \, d\theta \]
where \( k \) is Coulomb's constant.

### Step 5: Resolve the Electric Field into Components
The electric field \( dE \) has both vertical and horizontal components. Due to symmetry, the horizontal components will cancel out when integrated over the half ring. We only need to consider the vertical component:
\[ dE_y = dE \cos \theta = \frac{kq}{\pi R^2} \, d\theta \cos \theta \]

### Step 6: Integrate to Find the Total Electric Field
To find the total electric field \( E_y \) at the center, integrate \( dE_y \) from \( \theta = 0 \) to \( \theta = \pi \):
\[ E_y = \int_0^{\pi} dE_y = \int_0^{\pi} \frac{kq}{\pi R^2} \cos \theta \, d\theta \]

### Step 7: Calculate the Integral
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, since we are considering the vertical component, we need to consider only half of the integral (from \( 0 \) to \( \frac{\pi}{2} \)) and multiply by 2:
\[ E_y = 2 \int_0^{\frac{\pi}{2}} \frac{kq}{\pi R^2} \cos \theta \, d\theta = \frac{2kq}{\pi R^2} \left[ \sin \theta \right]_0^{\frac{\pi}{2}} = \frac{2kq}{\pi R^2} (1 - 0) = \frac{2kq}{\pi R^2} \]

### Step 8: Express in Terms of \( \epsilon_0 \)
Using \( k = \frac{1}{4 \pi \epsilon_0} \):
\[ E_y = \frac{2 \cdot \frac{1}{4 \pi \epsilon_0} q}{\pi R^2} = \frac{q}{2 \epsilon_0 R^2} \]

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

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