A circular ring of radius R and uniform linear charge density `+lamdaC//m` are kept in x - y plane with its centre at the origin. The electric field at a point `(0,0,R/sqrt(2))` is

Options

A. `lamda/(3sqrt(3)epsilon_0R)`
B. `(2lamda)/(3sqrt(3)epsilon_0R^2)`
C. `2/(3sqrt(3))lamda/(epsilon_0R)`
D. none the these

Answer

To find the electric field at the point \( (0, 0, \frac{R}{\sqrt{2}}) \) due to a circular ring of radius \( R \) with uniform linear charge density \( \lambda \), we can follow these steps:

### Step 1: Understand the Geometry
The circular ring lies in the x-y plane, centered at the origin. The point where we want to calculate the electric field is located on the z-axis at a distance \( \frac{R}{\sqrt{2}} \) above the center of the ring.

### Step 2: Use the Formula for Electric Field on the Axis of a Ring
The electric field \( E \) at a point along the axis of a charged ring can be calculated using the formula:

\[ E = \frac{k \cdot Q \cdot z}{(z^2 + R^2)^{3/2}} \]

where:
- \( k = \frac{1}{4\pi \epsilon_0} \) is Coulomb's constant,
- \( Q \) is the total charge on the ring,
- \( z \) is the distance from the center of the ring to the point along the axis,
- \( R \) is the radius of the ring.

### Step 3: Calculate the Total Charge \( Q \)
The total charge \( Q \) on the ring can be calculated as:

\[ Q = \lambda \cdot L \]

where \( L \) is the circumference of the ring, given by \( L = 2\pi R \). Thus,

\[ Q = \lambda \cdot 2\pi R \]

### Step 4: Substitute Values into the Electric Field Formula
Substituting \( Q \) into the electric field formula, we have:

\[ E = \frac{k \cdot (\lambda \cdot 2\pi R) \cdot \left(\frac{R}{\sqrt{2}}\right)}{\left(\left(\frac{R}{\sqrt{2}}\right)^2 + R^2\right)^{3/2}} \]

### Step 5: Simplify the Denominator
Calculating \( \left(\frac{R}{\sqrt{2}}\right)^2 + R^2 \):

\[ \left(\frac{R}{\sqrt{2}}\right)^2 = \frac{R^2}{2} \]

Thus,

\[ \frac{R^2}{2} + R^2 = \frac{R^2}{2} + \frac{2R^2}{2} = \frac{3R^2}{2} \]

Now, substituting this back into the electric field equation:

\[ E = \frac{k \cdot (\lambda \cdot 2\pi R) \cdot \left(\frac{R}{\sqrt{2}}\right)}{\left(\frac{3R^2}{2}\right)^{3/2}} \]

### Step 6: Calculate the Denominator
Calculating \( \left(\frac{3R^2}{2}\right)^{3/2} \):

\[ \left(\frac{3R^2}{2}\right)^{3/2} = \left(\frac{3^{3/2} R^3}{2^{3/2}}\right) = \frac{3\sqrt{3} R^3}{2\sqrt{2}} \]

### Step 7: Substitute Back and Simplify
Now substituting this back into the electric field expression:

\[ E = \frac{k \cdot (\lambda \cdot 2\pi R) \cdot \left(\frac{R}{\sqrt{2}}\right)}{\frac{3\sqrt{3} R^3}{2\sqrt{2}}} \]

This simplifies to:

\[ E = \frac{2k \lambda \pi R^2}{\frac{3\sqrt{3} R^3}{2\sqrt{2}}} \cdot \frac{1}{\sqrt{2}} \]

### Step 8: Final Simplification
After simplification, we find:

\[ E = \frac{4k \lambda \pi \sqrt{2}}{3\sqrt{3} R} \]

### Step 9: Substitute \( k \)
Substituting \( k = \frac{1}{4\pi \epsilon_0} \):

\[ E = \frac{4 \cdot \frac{1}{4\pi \epsilon_0} \cdot \lambda \cdot \pi \sqrt{2}}{3\sqrt{3} R} \]

This results in:

\[ E = \frac{\lambda \sqrt{2}}{3\sqrt{3} \epsilon_0 R} \]

### Final Answer
Thus, the electric field at the point \( (0, 0, \frac{R}{\sqrt{2}}) \) is:

\[ E = \frac{\lambda \sqrt{2}}{3\sqrt{3} \epsilon_0 R} \]