The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be

Options

A. `(1)/(4piepsilon_(0))(q)/(3sqrt(3)R^(2))`
B. `(1)/(4piepsilon_(0))(2q)/(3R^(2))`
C. `(1)/(4piepsilon_(0))(2q)/(3sqrt(3)R^(2))`
D. `(1)/(4piepsilon_(0))(3q)/(2sqrt(2)R^(2))`

Answer

To find the maximum electric field intensity on the axis of a uniformly charged ring of charge \( q \) and radius \( R \), we can follow these steps:

### Step 1: Understand the Electric Field due to a Ring
The electric field \( E \) at a point on the axis of a uniformly charged ring can be expressed as:
\[ E = \frac{k \cdot q \cdot x}{(x^2 + R^2)^{3/2}} \]
where:
- \( k \) is Coulomb's constant (\( k = \frac{1}{4 \pi \epsilon_0} \)),
- \( q \) is the total charge on the ring,
- \( R \) is the radius of the ring,
- \( x \) is the distance from the center of the ring along the axis.

### Step 2: Find the Condition for Maximum Electric Field
To find the maximum electric field, we need to maximize the expression for \( E \). This requires taking the derivative of \( E \) with respect to \( x \) and setting it to zero:
\[ \frac{dE}{dx} = 0 \]

### Step 3: Differentiate the Electric Field Expression
Let:
\[ u = \frac{x}{(x^2 + R^2)^{3/2}} \]
Then, we can express \( E \) as:
\[ E = k \cdot q \cdot u \]
To find the maximum of \( u \), we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = \frac{(x^2 + R^2)^{3/2} \cdot 1 - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2} \cdot 2x}{(x^2 + R^2)^3} \]
Setting \( \frac{du}{dx} = 0 \) gives:
\[ (x^2 + R^2)^{3/2} - 3x^2(x^2 + R^2)^{1/2} = 0 \]

### Step 4: Solve for \( x \)
Factoring out \( (x^2 + R^2)^{1/2} \):
\[ (x^2 + R^2)^{1/2} \left( (x^2 + R^2) - 3x^2 \right) = 0 \]
This simplifies to:
\[ R^2 - 2x^2 = 0 \implies x^2 = \frac{R^2}{2} \implies x = \frac{R}{\sqrt{2}} \]

### Step 5: Substitute \( x \) Back to Find Maximum Electric Field
Now, substitute \( x = \frac{R}{\sqrt{2}} \) back into the expression for \( E \):
\[ E_{\text{max}} = \frac{k \cdot q \cdot \frac{R}{\sqrt{2}}}{\left(\left(\frac{R}{\sqrt{2}}\right)^2 + R^2\right)^{3/2}} \]
Calculating the denominator:
\[ \left(\frac{R^2}{2} + R^2\right)^{3/2} = \left(\frac{3R^2}{2}\right)^{3/2} = \left(\frac{3^{3/2} R^3}{2^{3/2}}\right) \]
Thus:
\[ E_{\text{max}} = \frac{k \cdot q \cdot \frac{R}{\sqrt{2}}}{\frac{3\sqrt{3}R^3}{2\sqrt{2}}} = \frac{2kq}{3\sqrt{3}R^2} \]

### Step 6: Final Expression
Substituting \( k = \frac{1}{4\pi \epsilon_0} \):
\[ E_{\text{max}} = \frac{2q}{3\sqrt{3} \cdot 4\pi \epsilon_0 R^2} \]

### Conclusion
The maximum electric field intensity on the axis of a uniformly charged ring of charge \( q \) and radius \( R \) is given by:
\[ E_{\text{max}} = \frac{2q}{3\sqrt{3} \cdot 4\pi \epsilon_0 R^2} \]