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

### Step 1: Understand the Geometry
The circular ring lies in the x-y plane with its center at the origin. The point where we need to calculate the electric field is along the z-axis at a height of \(R/\sqrt{2}\).

### Step 2: Use the Electric Field Formula for a Ring
The electric field \(E\) at a point along the axis of a charged ring can be calculated using the formula:
\[ E = \frac{kQz}{(R^2 + z^2)^{3/2}} \]
where:
- \(k\) is Coulomb's constant (\(k = \frac{1}{4\pi \epsilon_0}\)),
- \(Q\) 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 expressed in terms of the linear charge density \(\lambda\):
\[ Q = \lambda \cdot (2\pi R) \]
This is because the circumference of the ring is \(2\pi R\).

### Step 4: Substitute Values into the Electric Field Formula
Now, substituting \(Q\) into the electric field formula:
\[ E = \frac{k(\lambda \cdot 2\pi R)z}{(R^2 + z^2)^{3/2}} \]
Here, \(z = \frac{R}{\sqrt{2}}\).

### Step 5: Substitute \(z\) into the Formula
Substituting \(z\) into the equation:
\[ E = \frac{k(\lambda \cdot 2\pi R)(\frac{R}{\sqrt{2}})}{(R^2 + (\frac{R}{\sqrt{2}})^2)^{3/2}} \]

### Step 6: Simplify the Denominator
Calculate \(R^2 + \left(\frac{R}{\sqrt{2}}\right)^2\):
\[ R^2 + \frac{R^2}{2} = \frac{2R^2 + R^2}{2} = \frac{3R^2}{2} \]
Thus, the denominator becomes:
\[ \left(\frac{3R^2}{2}\right)^{3/2} = \left(\frac{3^{3/2} R^3}{2^{3/2}}\right) = \frac{3\sqrt{3} R^3}{2\sqrt{2}} \]

### Step 7: Substitute Denominator Back into the Electric Field Formula
Now substitute this back into the formula for \(E\):
\[ E = \frac{k(\lambda \cdot 2\pi R)(\frac{R}{\sqrt{2}})}{\frac{3\sqrt{3} R^3}{2\sqrt{2}}} \]

### Step 8: Simplify the Expression
Now simplify the expression:
\[ E = \frac{2k\lambda\pi R^2/\sqrt{2}}{3\sqrt{3} R^3/2\sqrt{2}} = \frac{2k\lambda\pi R^2}{3\sqrt{3} R^3} = \frac{2k\lambda\pi}{3\sqrt{3} R} \]

### Step 9: Substitute \(k\) with \(\epsilon_0\)
Finally, substituting \(k = \frac{1}{4\pi \epsilon_0}\):
\[ E = \frac{2(\frac{1}{4\pi \epsilon_0})\lambda\pi}{3\sqrt{3} R} = \frac{\lambda}{6\sqrt{3}\epsilon_0 R} \]

### Final Result
Thus, the electric field at the point \((0, 0, R/\sqrt{2})\) is:
\[ E = \frac{\lambda}{6\sqrt{3}\epsilon_0 R} \]