A space charge of density `rho` is uniformly distributed in an infinitely long cylinder of radius R Then, for any point at distance r from the axis (relative permittivity = 1)

Options

A. the electric field is `E=(rhor)/(2 in_0)` for 0 `lt` r `lt` R
B. the electric field is `E=(rhoR^3)/(2in_0r^2)` for r `gt` R
C. the electric field is `E=(rhor)/in_0` for 0 `lt` r `lt` R
D. the electric field is `E=(rhoR^2)/(in_0r)` for r `gt` R

Answer

To solve the problem of finding the electric field inside and outside an infinitely long cylinder with a uniform space charge density \(\rho\), we will apply Gauss's law.

### Step-by-Step Solution:

1. **Identify the Geometry and Charge Distribution**:
- We have an infinitely long cylinder with radius \(R\) and a uniform charge density \(\rho\). We need to find the electric field at a distance \(r\) from the axis of the cylinder.

2. **Consider the Case Inside the Cylinder (\(r < R\))**:
- For points inside the cylinder, we will use a Gaussian surface in the form of a cylinder of radius \(r\) (where \(r < R\)) and length \(L\).
- The electric field \(E\) is assumed to be uniform and radial due to symmetry.

3. **Apply Gauss's Law**:
- According to Gauss's law, the electric flux through the Gaussian surface is equal to the charge enclosed divided by \(\epsilon_0\):
\[ \Phi_E = E \cdot A = E \cdot (2\pi r L) \]
- The charge enclosed \(Q_{\text{enc}}\) within the Gaussian surface is given by:
\[ Q_{\text{enc}} = \rho \cdot V = \rho \cdot (\pi r^2 L) \]
- Thus, Gauss's law gives us:
\[ E \cdot (2\pi r L) = \frac{\rho \cdot (\pi r^2 L)}{\epsilon_0} \]

4. **Solve for the Electric Field Inside the Cylinder**:
- Cancel \(L\) and \(\pi\) from both sides:
\[ E \cdot (2r) = \frac{\rho r^2}{\epsilon_0} \]
- Rearranging gives:
\[ E = \frac{\rho r}{2 \epsilon_0} \]

5. **Consider the Case Outside the Cylinder (\(r > R\))**:
- For points outside the cylinder, we will again use a Gaussian surface in the form of a cylinder of radius \(r\) (where \(r > R\)).
- The charge enclosed is the total charge of the cylinder up to radius \(R\):
\[ Q_{\text{enc}} = \rho \cdot (\pi R^2 L) \]
- Using Gauss's law, we have:
\[ E \cdot (2\pi r L) = \frac{\rho \cdot (\pi R^2 L)}{\epsilon_0} \]

6. **Solve for the Electric Field Outside the Cylinder**:
- Cancel \(L\) and \(\pi\) from both sides:
\[ E \cdot (2r) = \frac{\rho R^2}{\epsilon_0} \]
- Rearranging gives:
\[ E = \frac{\rho R^2}{2 \epsilon_0 r} \]

### Summary of Results:
- **Electric Field Inside the Cylinder (\(r < R\))**:
\[ E = \frac{\rho r}{2 \epsilon_0} \]
- **Electric Field Outside the Cylinder (\(r > R\))**:
\[ E = \frac{\rho R^2}{2 \epsilon_0 r} \]