Graph shows magnitude of electric field `(E)`, charge enclosed `(q_("enclosed:))` within the concentic sphere and net flux `(phi)` through a concentric spherical Gaussian surface as a function of distance (r) from centre of a uniformly positive charged solid sphere of radius R. Choose the correct option (s):

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the relationships between the electric field \( E \), the charge enclosed \( q_{\text{enclosed}} \), and the net electric flux \( \Phi \) through a concentric spherical Gaussian surface surrounding a uniformly positively charged solid sphere of radius \( R \).

### Step-by-Step Solution:

1. **Understanding Charge Density**:
- Let \( \rho \) be the volume charge density of the uniformly charged sphere. The total charge \( Q \) of the sphere can be expressed as:
\[ Q = \rho \cdot \frac{4}{3} \pi R^3 \]

2. **Electric Field Inside the Sphere**:
- For a point inside the sphere (where \( r < R \)), the charge enclosed \( q_{\text{enclosed}} \) within a Gaussian surface of radius \( r \) is:
\[ q_{\text{enclosed}} = \rho \cdot \frac{4}{3} \pi r^3 \]
- According to Gauss's law, the electric field \( E \) at a distance \( r \) from the center is given by:
\[ E \cdot 4 \pi r^2 = \frac{q_{\text{enclosed}}}{\epsilon_0} \]
- Substituting for \( q_{\text{enclosed}} \):
\[ E \cdot 4 \pi r^2 = \frac{\rho \cdot \frac{4}{3} \pi r^3}{\epsilon_0} \]
- Simplifying gives:
\[ E = \frac{\rho r}{3 \epsilon_0} \]

3. **Electric Field Outside the Sphere**:
- For a point outside the sphere (where \( r \geq R \)), the charge enclosed is simply the total charge \( Q \):
\[ q_{\text{enclosed}} = Q = \rho \cdot \frac{4}{3} \pi R^3 \]
- Again applying Gauss's law:
\[ E \cdot 4 \pi r^2 = \frac{Q}{\epsilon_0} \]
- Substituting for \( Q \):
\[ E \cdot 4 \pi r^2 = \frac{\rho \cdot \frac{4}{3} \pi R^3}{\epsilon_0} \]
- This simplifies to:
\[ E = \frac{Q}{4 \pi \epsilon_0 r^2} = \frac{\rho \cdot \frac{4}{3} \pi R^3}{4 \pi \epsilon_0 r^2} = \frac{\rho R^3}{3 \epsilon_0 r^2} \]

4. **Net Electric Flux**:
- The net electric flux \( \Phi \) through the Gaussian surface is given by:
\[ \Phi = \frac{q_{\text{enclosed}}}{\epsilon_0} \]
- For \( r < R \):
\[ \Phi = \frac{\rho \cdot \frac{4}{3} \pi r^3}{\epsilon_0} \]
- For \( r \geq R \):
\[ \Phi = \frac{\rho \cdot \frac{4}{3} \pi R^3}{\epsilon_0} \]

5. **Graph Analysis**:
- The electric field \( E \) increases linearly with \( r \) for \( r < R \) and decreases with \( \frac{1}{r^2} \) for \( r \geq R \).
- The charge enclosed \( q_{\text{enclosed}} \) increases with \( r^3 \) for \( r < R \) and remains constant for \( r \geq R \).
- The net electric flux \( \Phi \) is constant for \( r \geq R \) and increases with \( r^3 \) for \( r < R \).

### Conclusion:
Based on the analysis, the correct options for the graphs representing \( E \), \( q_{\text{enclosed}} \), and \( \Phi \) as functions of \( r \) are:
- **Electric Field \( E \)**: Increases linearly for \( r < R \) and decreases with \( \frac{1}{r^2} \) for \( r \geq R \).
- **Charge Enclosed \( q_{\text{enclosed}} \)**: Increases with \( r^3 \) for \( r < R \) and remains constant for \( r \geq R \).
- **Net Electric Flux \( \Phi \)**: Increases with \( r^3 \) for \( r < R \) and remains constant for \( r \geq R \).