A system consits fo a thin charged wire ring of radius `R` and a very long uniformly charged thread oriented along the axis of the ring, with one of its ends coinciding with the centre of the ring. The total charge of the ring, with one of the ring so equal to `q`. The charge of the thread (per unit length) is equal to `lambda`. Find the interaction froce between the ring and the thread.

Options

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

Explanation

To find the interaction force between a charged ring and a uniformly charged thread, we can follow these steps:

### Step 1: Understand the System
We have:
- A thin charged wire ring of radius \( R \) with total charge \( q \).
- A very long uniformly charged thread with linear charge density \( \lambda \), extending along the axis of the ring, with one end at the center of the ring.

### Step 2: Electric Field Due to the Ring
The electric field \( E \) at a distance \( x \) from the center of the ring along its axis can be derived using the formula:
\[ E = \frac{k \cdot q \cdot x}{(R^2 + x^2)^{3/2}} \]
where \( k = \frac{1}{4\pi\epsilon_0} \).

### Step 3: Charge Element of the Thread
Consider a small element of the thread of length \( dx \) at a distance \( x \) from the center of the ring. The charge on this small element is given by:
\[ dq = \lambda \cdot dx \]

### Step 4: Force on the Charge Element
The force \( dF \) acting on this small charge element \( dq \) due to the electric field \( E \) produced by the ring is:
\[ dF = dq \cdot E = \lambda \cdot dx \cdot \frac{k \cdot q \cdot x}{(R^2 + x^2)^{3/2}} \]

### Step 5: Total Force on the Thread
To find the total force \( F \) acting on the thread, we integrate \( dF \) from \( x = 0 \) to \( x = \infty \):
\[ F = \int_0^{\infty} dF = \int_0^{\infty} \frac{k \cdot q \cdot \lambda \cdot x \cdot dx}{(R^2 + x^2)^{3/2}} \]

### Step 6: Substituting Variables
Let \( t^2 = R^2 + x^2 \), then \( 2x \, dx = 2t \, dt \) or \( x \, dx = t \, dt \). The limits change from \( x = 0 \) (where \( t = R \)) to \( x = \infty \) (where \( t = \infty \)):
\[ F = \frac{k \cdot q \cdot \lambda}{4\pi\epsilon_0} \int_R^{\infty} \frac{t \, dt}{t^3} = \frac{k \cdot q \cdot \lambda}{4\pi\epsilon_0} \int_R^{\infty} \frac{1}{t^2} dt \]

### Step 7: Evaluate the Integral
The integral \( \int \frac{1}{t^2} dt = -\frac{1}{t} \):
\[ F = \frac{k \cdot q \cdot \lambda}{4\pi\epsilon_0} \left[-\frac{1}{t}\right]_R^{\infty} = \frac{k \cdot q \cdot \lambda}{4\pi\epsilon_0} \left(0 + \frac{1}{R}\right) \]
Thus,
\[ F = \frac{k \cdot q \cdot \lambda}{4\pi\epsilon_0 R} \]

### Step 8: Final Result
Substituting \( k = \frac{1}{4\pi\epsilon_0} \):
\[ F = \frac{q \cdot \lambda}{4\pi\epsilon_0 R} \]
This is the interaction force between the ring and the thread.

### Summary
The interaction force between the charged ring and the uniformly charged thread is given by:
\[ F = \frac{q \cdot \lambda}{4\pi\epsilon_0 R} \]