Electric field at centre of quarter circular ring having charge density `lambda` is _____

Options

A. `(sqrt(2)lambda)/(4piepsilon_(0)r)`
B. `(lambda)/(2piepsilon_(0)r)`
C. `(lambda)/(epsilon_(0)r)`
D. `(sqrt(2)lambda)/(2piepsilon_(0)r)`

Answer

To find the electric field at the center of a quarter circular ring with a linear charge density \( \lambda \), we can follow these steps:

### Step 1: Understand the Geometry
We have a quarter circular ring of radius \( r \) in the xy-plane, centered at the origin. The charge density is given as \( \lambda \).

### Step 2: Consider a Small Element of Charge
Let’s consider a small segment of the ring, \( dq \), which subtends an angle \( d\theta \) at the center. The length of this small segment is \( r \, d\theta \).

### Step 3: Express the Charge Element
The charge \( dq \) can be expressed in terms of the linear charge density:
\[ dq = \lambda \cdot (r \, d\theta) \]

### Step 4: Calculate the Electric Field Contribution
The electric field \( dE \) due to the charge element \( dq \) at the center of the ring is given by Coulomb's law:
\[ dE = \frac{k \, dq}{r^2} \]
Substituting \( dq \):
\[ dE = \frac{k \, \lambda \, (r \, d\theta)}{r^2} = \frac{k \, \lambda}{r} \, d\theta \]

### Step 5: Determine the Direction of the Electric Field
The electric field \( dE \) has both x and y components. The angle \( \theta \) gives the direction:
- The x-component is \( dE_x = dE \cos(\theta) \)
- The y-component is \( dE_y = dE \sin(\theta) \)

Thus, we can write:
\[ dE_x = \frac{k \lambda}{r} \cos(\theta) d\theta \]
\[ dE_y = \frac{k \lambda}{r} \sin(\theta) d\theta \]

### Step 6: Integrate to Find Total Electric Field
To find the total electric field \( E \), we integrate \( dE_x \) and \( dE_y \) from \( 0 \) to \( \frac{\pi}{2} \):
\[ E_x = \int_0^{\frac{\pi}{2}} dE_x = \int_0^{\frac{\pi}{2}} \frac{k \lambda}{r} \cos(\theta) d\theta \]
\[ E_y = \int_0^{\frac{\pi}{2}} dE_y = \int_0^{\frac{\pi}{2}} \frac{k \lambda}{r} \sin(\theta) d\theta \]

### Step 7: Evaluate the Integrals
The integrals evaluate to:
\[ E_x = \frac{k \lambda}{r} [\sin(\theta)]_0^{\frac{\pi}{2}} = \frac{k \lambda}{r} (1 - 0) = \frac{k \lambda}{r} \]
\[ E_y = \frac{k \lambda}{r} [-\cos(\theta)]_0^{\frac{\pi}{2}} = \frac{k \lambda}{r} (0 - (-1)) = \frac{k \lambda}{r} \]

### Step 8: Combine the Components
The total electric field \( E \) at the center is:
\[ E = E_x \hat{i} + E_y \hat{j} = \frac{k \lambda}{r} \hat{i} + \frac{k \lambda}{r} \hat{j} \]
This can be expressed as:
\[ E = \frac{k \lambda}{r} (\hat{i} + \hat{j}) \]

### Step 9: Calculate the Magnitude
The magnitude of the electric field is:
\[ |E| = \frac{k \lambda}{r} \sqrt{1^2 + 1^2} = \frac{k \lambda}{r} \sqrt{2} \]

### Step 10: Substitute \( k \)
Using \( k = \frac{1}{4\pi \epsilon_0} \):
\[ |E| = \frac{\sqrt{2} \lambda}{4 \pi \epsilon_0 r} \]

### Final Answer
The electric field at the center of the quarter circular ring is:
\[ E = \frac{\sqrt{2} \lambda}{4 \pi \epsilon_0 r} \]