For a uniformly charged non conducting sphere of radius R which of following shows a correct graph between the electric field intensity and the distance from the centre of sphere –

Options

A.
B.
C.
D.

Explanation

To solve the problem of determining the correct graph between electric field intensity and distance from the center of a uniformly charged non-conducting sphere, we will analyze the electric field both inside and outside the sphere step by step.

### Step 1: Understanding the Problem
We need to find the relationship between the electric field intensity (E) and the distance (r) from the center of a uniformly charged non-conducting sphere of radius R.

### Step 2: Electric Field Inside the Sphere
1. **Define the Gaussian Surface**: For a point inside the sphere (r < R), we consider a Gaussian surface that is a sphere of radius r, concentric with the charged sphere.

2. **Calculate the Charge Inside**: The charge enclosed by this Gaussian surface can be calculated using the volume charge density. If the total charge of the sphere is Q, the charge inside the Gaussian surface is given by:
\[ Q_{\text{inside}} = Q \cdot \frac{V_{\text{small sphere}}}{V_{\text{large sphere}}} = Q \cdot \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = Q \cdot \frac{r^3}{R^3} \]

3. **Apply Gauss's Law**: According to Gauss's law, the electric flux through the Gaussian surface is:
\[ \Phi = E \cdot 4\pi r^2 = \frac{Q_{\text{inside}}}{\epsilon_0} \]
Substituting for \(Q_{\text{inside}}\):
\[ E \cdot 4\pi r^2 = \frac{Q \cdot \frac{r^3}{R^3}}{\epsilon_0} \]

4. **Solve for Electric Field (E)**: Rearranging gives:
\[ E = \frac{Q}{4\pi \epsilon_0 R^3} r \]
This shows that the electric field inside the sphere varies linearly with r.

### Step 3: Electric Field Outside the Sphere
1. **Define the Gaussian Surface**: For a point outside the sphere (r > R), we consider a Gaussian surface that is a sphere of radius r.

2. **Apply Gauss's Law**: The total charge enclosed is still Q. Thus, the electric flux through the Gaussian surface is:
\[ \Phi = E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0} \]

3. **Solve for Electric Field (E)**: Rearranging gives:
\[ E = \frac{Q}{4\pi \epsilon_0 r^2} \]
This indicates that the electric field outside the sphere varies inversely with the square of the distance from the center.

### Step 4: Summary of Results
- **Inside the Sphere**: \(E \propto r\) (Electric field increases linearly with distance)
- **Outside the Sphere**: \(E \propto \frac{1}{r^2}\) (Electric field decreases with the square of the distance)

### Step 5: Graph Interpretation
- The graph of electric field intensity vs. distance from the center will show a linear increase for distances less than R and a decrease proportional to \(1/r^2\) for distances greater than R.

### Conclusion
The correct graph will start from the origin, increase linearly until it reaches the radius R, and then drop off as \(1/r^2\) for distances greater than R.