A point charge `q` is located at the centre fo a thin ring of radius `R` with uniformly distributed charge `-q`, find the magnitude of the electric field strength vectro at the point lying on the axis of the ring at a distance `x` from its centre, if `x gt gt R`.

Options

A. `(QR^(2))/(8pi epsilon_(0)x^(4))`
B. `(3QR^(2))/(8 pi epsilon_(0)x^(4))`
C. `(5QR^(2))/(8pi epsilon_(0)x^(4))`
D. `(7QR^(2))/(8pi epsilon_(0)x^(4))`

Explanation

To find the magnitude of the electric field strength vector at a point lying on the axis of a thin ring with a point charge at its center, we can follow these steps:

### Step 1: Understand the Configuration
We have a point charge \( q \) located at the center of a thin ring of radius \( R \) that has a uniformly distributed charge of \( -q \). We need to find the electric field at a point \( P \) on the axis of the ring at a distance \( x \) from its center, where \( x \gg R \).

### Step 2: Determine the Electric Field Due to the Point Charge
The electric field \( E_1 \) due to the point charge \( q \) at point \( P \) can be calculated using Coulomb's law:
\[ E_1 = \frac{kq}{x^2} \]
where \( k \) is Coulomb's constant.

### Step 3: Determine the Electric Field Due to the Ring Charge
The electric field \( E_2 \) due to the uniformly charged ring can be derived from the principle of superposition. The electric field at point \( P \) due to the ring is directed along the axis of the ring and can be expressed as:
\[ E_2 = -\frac{k(-q)x}{(x^2 + R^2)^{3/2}} \]
Here, the negative sign indicates that the electric field due to the negative charge is directed towards the ring.

### Step 4: Combine the Electric Fields
The net electric field \( E \) at point \( P \) is the vector sum of \( E_1 \) and \( E_2 \):
\[ E = E_1 + E_2 = \frac{kq}{x^2} - \frac{k(-q)x}{(x^2 + R^2)^{3/2}} \]

### Step 5: Simplify the Expression
Since \( x \gg R \), we can approximate \( x^2 + R^2 \approx x^2 \). Thus, we can simplify \( E_2 \):
\[ E_2 \approx -\frac{k(-q)x}{(x^2)^{3/2}} = -\frac{k(-q)x}{x^3} = \frac{kq}{x^2} \]
Now substituting this back into the expression for the net electric field:
\[ E \approx \frac{kq}{x^2} + \frac{kq}{x^2} = \frac{2kq}{x^2} \]

### Step 6: Final Expression
Thus, the magnitude of the electric field strength vector at point \( P \) is:
\[ E \approx \frac{2kq}{x^2} \]

### Step 7: Substitute Coulomb's Constant
If we substitute \( k = \frac{1}{4\pi \epsilon_0} \):
\[ E \approx \frac{2 \cdot \frac{1}{4\pi \epsilon_0} \cdot q}{x^2} = \frac{q}{2\pi \epsilon_0 x^2} \]

### Final Answer
The magnitude of the electric field strength vector at point \( P \) is:
\[ E = \frac{q}{2\pi \epsilon_0 x^2} \]