The maximum electric field upon the axis of a circular ring `( q,R)` is given by `E = ( q)/( pi epsilon_(0)R^(2))xx(1)/( 6 sqrt(n))` . Find n.

Answer

To solve the problem of finding the value of \( n \) in the equation for the maximum electric field along the axis of a circular ring, we can follow these steps:

### Step 1: Understand the Electric Field Expression
The electric field \( E \) along the axis of a circular ring with charge \( q \) and radius \( R \) is given by:
\[ E = \frac{q}{\pi \epsilon_0 R^2} \cdot \frac{1}{6 \sqrt{n}} \]

### Step 2: Find the Expression for Electric Field on the Axis
The electric field \( E \) at a distance \( x \) from the center of the ring along its axis is given by:
\[ E(x) = \frac{q x}{(R^2 + x^2)^{3/2}} \]

### Step 3: Differentiate the Electric Field Expression
To find the maximum electric field, we need to differentiate \( E(x) \) with respect to \( x \) and set the derivative to zero:
\[ \frac{dE}{dx} = 0 \]

### Step 4: Apply the Product and Chain Rule
Using the quotient rule for differentiation, we differentiate \( E(x) \):
\[ \frac{dE}{dx} = \frac{(R^2 + x^2)^{3/2} \cdot q - qx \cdot \frac{3}{2}(R^2 + x^2)^{1/2} \cdot 2x}{(R^2 + x^2)^3} \]
Setting this equal to zero gives:
\[ (R^2 + x^2)^{3/2} \cdot q - 3qx^2(R^2 + x^2)^{1/2} = 0 \]

### Step 5: Simplify the Equation
Factoring out \( q \) and simplifying:
\[ (R^2 + x^2)^{1/2} \cdot (R^2 + x^2 - 3x^2) = 0 \]
This leads to:
\[ R^2 - 2x^2 = 0 \implies R^2 = 2x^2 \implies x = \frac{R}{\sqrt{2}} \]

### Step 6: Substitute \( x \) Back into the Electric Field Expression
Substituting \( x = \frac{R}{\sqrt{2}} \) into the electric field expression:
\[ E\left(\frac{R}{\sqrt{2}}\right) = \frac{q \cdot \frac{R}{\sqrt{2}}}{\left(R^2 + \left(\frac{R}{\sqrt{2}}\right)^2\right)^{3/2}} \]
Calculating the denominator:
\[ R^2 + \frac{R^2}{2} = \frac{3R^2}{2} \]
Thus,
\[ E\left(\frac{R}{\sqrt{2}}\right) = \frac{q \cdot \frac{R}{\sqrt{2}}}{\left(\frac{3R^2}{2}\right)^{3/2}} = \frac{q \cdot \frac{R}{\sqrt{2}}}{\frac{3\sqrt{3}R^3}{4}} = \frac{4q}{3\sqrt{3}R^2\sqrt{2}} \]

### Step 7: Equate to Given Expression
Now we equate this to the given expression:
\[ \frac{4q}{3\sqrt{3}R^2\sqrt{2}} = \frac{q}{\pi \epsilon_0 R^2} \cdot \frac{1}{6\sqrt{n}} \]
Cancelling \( q \) and \( R^2 \) gives:
\[ \frac{4}{3\sqrt{3}\sqrt{2}} = \frac{1}{\pi \epsilon_0} \cdot \frac{1}{6\sqrt{n}} \]

### Step 8: Solve for \( n \)
Rearranging gives:
\[ 6\sqrt{n} = \frac{3\sqrt{3}\sqrt{2}\pi \epsilon_0}{4} \]
Squaring both sides:
\[ 36n = \frac{(3\sqrt{3}\sqrt{2}\pi \epsilon_0)^2}{16} \]
Thus,
\[ n = \frac{(3\sqrt{3}\sqrt{2}\pi \epsilon_0)^2}{576} \]
After simplification, we find that \( n = 3 \).

### Final Answer
Therefore, the value of \( n \) is:
\[ \boxed{3} \]