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

Options

A. `1/4πepsilon_0 q/3sqrt(2)R^2`
B. `1/4πepsilon_0 2q/3R^2`
C. `1/4πepsilon_0 q/3sqrt(3)R^2`
D. `1/4πepsilon_0 2q/3sqrt(3)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 Charged Ring
The electric field \( E \) at a point on the axis of a uniformly charged ring at a distance \( x \) from the center of the ring is given by the formula:

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

where:
- \( E \) is the electric field intensity,
- \( k = \frac{1}{4 \pi \epsilon_0} \) is Coulomb's constant,
- \( q \) is the total charge on the ring,
- \( R \) is the radius of the ring,
- \( x \) is the distance from the center of the ring to the point where the electric field is being calculated.

### Step 2: Differentiate the Electric Field Expression
To find the maximum electric field, we need to differentiate the expression for \( E \) with respect to \( x \) and set the derivative equal to zero:

\[ \frac{dE}{dx} = 0 \]

Using the quotient rule for differentiation, we differentiate \( E \):

\[ \frac{dE}{dx} = \frac{(x^2 + R^2)^{3/2} \cdot (k \cdot q) - k \cdot q \cdot x \cdot \frac{3}{2}(x^2 + R^2)^{1/2} \cdot 2x}{(x^2 + R^2)^3} \]

### Step 3: Set the Derivative to Zero
Setting the numerator of the derivative equal to zero gives us:

\[ (x^2 + R^2)^{3/2} - 3x^2(x^2 + R^2)^{1/2} = 0 \]

This simplifies to:

\[ (x^2 + R^2) = 3x^2 \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:

\[ R^2 = 2x^2 \implies x^2 = \frac{R^2}{2} \implies x = \frac{R}{\sqrt{2}} \]

### Step 5: Substitute \( x \) Back to Find Maximum Electric Field
Now we substitute \( x = \frac{R}{\sqrt{2}} \) back into the electric field equation to find \( E_{\text{max}} \):

\[ E_{\text{max}} = \frac{k \cdot q \cdot \frac{R}{\sqrt{2}}}{\left(\left(\frac{R}{\sqrt{2}}\right)^2 + R^2\right)^{3/2}} \]

Calculating the denominator:

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

Now substituting back into the equation for \( E_{\text{max}} \):

\[ E_{\text{max}} = \frac{k \cdot q \cdot \frac{R}{\sqrt{2}}}{\frac{3\sqrt{3} R^3}{2\sqrt{2}}} \]

### Step 6: Simplify the Expression
Simplifying gives:

\[ E_{\text{max}} = \frac{2kq}{3\sqrt{3}R^2} \]

### Final Result
Thus, the maximum electric field intensity on the axis of a uniformly charged ring is:

\[ E_{\text{max}} = \frac{2kq}{3\sqrt{3}R^2} \]