Charge is uniformly distributed with density `rho` in a cylindrical region of radius R and infinite length. The magnitude of electric field at distance `(R )/(2)` and 2R from the axis of the cylinder is `E_(1)` and `E_(2)` respectively. The ratio `(E_(1))/(E_(2))` is equal to ___________.

Explanation

To solve the problem of finding the ratio \((E_1)/(E_2)\) for the electric field at distances \((R/2)\) and \(2R\) from the axis of a uniformly charged infinite cylinder, we can use Gauss's law. Here's a step-by-step solution:

### Step 1: Understanding the Problem
We have a uniformly charged infinite cylinder with charge density \(\rho\) and radius \(R\). We need to find the electric field at two different distances from the axis of the cylinder: \(E_1\) at \(r = R/2\) and \(E_2\) at \(r = 2R\).

### Step 2: Applying Gauss's Law
Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (\(\epsilon_0\)):
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Where \(\Phi_E = \oint \vec{E} \cdot d\vec{A}\).

### Step 3: Finding \(E_1\) at \(r = R/2\)
1. **Choose a Gaussian surface**: A cylindrical surface of radius \(R/2\) and length \(L\).
2. **Calculate the enclosed charge**: The volume of the cylinder of radius \(R/2\) and length \(L\) is:
\[ V = \pi \left(\frac{R}{2}\right)^2 L = \frac{\pi R^2 L}{4} \]
The enclosed charge \(Q_{\text{enc}}\) is:
\[ Q_{\text{enc}} = \rho \cdot V = \rho \cdot \frac{\pi R^2 L}{4} \]
3. **Calculate the electric field**: The electric field \(E_1\) is uniform over the curved surface of the cylinder:
\[ \Phi_E = E_1 \cdot (2\pi \cdot \frac{R}{2} \cdot L) = E_1 \cdot \pi R L \]
Setting the two expressions for flux equal gives:
\[ E_1 \cdot \pi R L = \frac{\rho \cdot \frac{\pi R^2 L}{4}}{\epsilon_0} \]
Simplifying:
\[ E_1 = \frac{\rho R}{4 \epsilon_0} \]

### Step 4: Finding \(E_2\) at \(r = 2R\)
1. **Choose a Gaussian surface**: A cylindrical surface of radius \(2R\) and length \(L\).
2. **Calculate the enclosed charge**: The entire charge within the cylinder of radius \(R\) is:
\[ Q_{\text{enc}} = \rho \cdot \pi R^2 L \]
3. **Calculate the electric field**: The electric field \(E_2\) is uniform over the curved surface:
\[ \Phi_E = E_2 \cdot (2\pi \cdot 2R \cdot L) = E_2 \cdot 4\pi R L \]
Setting the two expressions for flux equal gives:
\[ E_2 \cdot 4\pi R L = \frac{\rho \cdot \pi R^2 L}{\epsilon_0} \]
Simplifying:
\[ E_2 = \frac{\rho R}{4 \epsilon_0} \]

### Step 5: Finding the Ratio \((E_1)/(E_2)\)
Now we can find the ratio:
\[ \frac{E_1}{E_2} = \frac{\frac{\rho R}{4 \epsilon_0}}{\frac{\rho R}{4 \epsilon_0}} = 1 \]

### Final Answer
The ratio \(\frac{E_1}{E_2}\) is equal to **1**.