A ring of radius `a` contains a charge `q` distributed uniformly over its length. Find the electric field at a point on the axis of the ring at a distance `x` from the centre.

Options

A. `(kq^2)/(x(a^2+x^2)^(3/2))`
B. `(2kq^2x)/(a^2+x^2)^(3/2)`
C. `(kq^2x)/(a^2+x^2)^(3/2)`
D. `(kq^2)/(2x(a^2+x^2)^(3/2))`

Explanation

To find the electric field at a point on the axis of a uniformly charged ring, we can follow these steps:

### Step 1: Understand the Geometry
Consider a ring of radius \( a \) with a total charge \( q \) uniformly distributed along its circumference. We want to find the electric field at a point located on the axis of the ring at a distance \( x \) from the center of the ring.

### Step 2: Define Charge Element
Let’s consider a small charge element \( dq \) on the ring. The total charge \( q \) is uniformly distributed, so we can express \( dq \) in terms of the linear charge density \( \lambda \):
\[ \lambda = \frac{q}{2\pi a} \]
Thus, for a small arc length \( ds \) on the ring, we have:
\[ dq = \lambda \, ds = \frac{q}{2\pi a} \, ds \]

### Step 3: Electric Field Due to Charge Element
The electric field \( dE \) due to the charge element \( dq \) at the point on the axis can be expressed in terms of its components. The electric field \( dE \) has two components: one along the axis (let’s call it \( dE_x \)) and one perpendicular to the axis (let’s call it \( dE_y \)). Due to symmetry, the \( dE_y \) components from opposite sides of the ring will cancel out, leaving only the \( dE_x \) component.

The distance from the charge element to the point on the axis is:
\[ r = \sqrt{x^2 + a^2} \]

The \( dE \) due to \( dq \) is given by Coulomb's law:
\[ dE = \frac{k \, dq}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 4: Resolve Electric Field Components
The component of the electric field along the axis \( dE_x \) is:
\[ dE_x = dE \cdot \cos(\theta) \]
where \( \cos(\theta) = \frac{x}{\sqrt{x^2 + a^2}} \). Therefore,
\[ dE_x = \frac{k \, dq}{r^2} \cdot \frac{x}{\sqrt{x^2 + a^2}} = \frac{k \, dq \cdot x}{(x^2 + a^2)^{3/2}} \]

### Step 5: Integrate Over the Ring
Now we need to integrate \( dE_x \) over the entire ring. The total electric field \( E \) at the point on the axis is:
\[ E = \int dE_x = \int \frac{k \, dq \cdot x}{(x^2 + a^2)^{3/2}} \]
Since \( dq = \frac{q}{2\pi a} \, ds \) and \( ds = a \, d\theta \) (where \( \theta \) is the angle parameterizing the ring), we have:
\[ dq = \frac{q}{2\pi} \, d\theta \]
Thus,
\[ E = \int_0^{2\pi} \frac{k \cdot \frac{q}{2\pi} \cdot x}{(x^2 + a^2)^{3/2}} \, d\theta \]
The integral simplifies to:
\[ E = \frac{k \cdot q \cdot x}{(x^2 + a^2)^{3/2}} \cdot \int_0^{2\pi} \frac{1}{2\pi} \, d\theta = \frac{k \cdot q \cdot x}{(x^2 + a^2)^{3/2}} \]

### Step 6: Final Expression
Thus, the electric field at a distance \( x \) from the center of the ring along the axis is:
\[ E = \frac{k \cdot q \cdot x}{(x^2 + a^2)^{3/2}} \]