A spherical insulator of radius R is charged uniformly with a charge Q throughout its volume and contains a point charge `Q/16` located at its centre. Which of the following graphs best represents qualitatively, the variation of electric field intensity E with distance r from the centre?

Options

A.
B.
C.
D.

Explanation

To solve the problem of determining the variation of electric field intensity \( E \) with distance \( r \) from the center of a uniformly charged spherical insulator containing a point charge at its center, we can follow these steps:

### Step 1: Understand the Setup
We have a spherical insulator of radius \( R \) uniformly charged with a total charge \( Q \) and a point charge \( \frac{Q}{16} \) located at its center. We need to analyze how the electric field \( E \) behaves at different distances \( r \) from the center.

### Step 2: Electric Field Inside the Sphere
For a point inside the uniformly charged sphere (i.e., \( r < R \)), the electric field due to the uniformly charged sphere can be calculated using Gauss's law. The charge enclosed within a Gaussian surface of radius \( r \) is given by:

\[ Q_{\text{enc}} = \frac{Q}{V} \times \text{Volume of sphere of radius } r = \frac{Q}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi r^3 = Q \frac{r^3}{R^3} \]

The electric field \( E \) at a distance \( r \) from the center is given by:

\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q_{\text{enc}}}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{Q \frac{r^3}{R^3}}{r^2} = \frac{Q}{4 \pi \epsilon_0 R^3} r \]

Additionally, we must consider the electric field due to the point charge \( \frac{Q}{16} \):

\[ E_{\text{point}} = \frac{1}{4 \pi \epsilon_0} \frac{\frac{Q}{16}}{r^2} \]

Thus, the total electric field \( E \) inside the sphere (for \( r < R \)) becomes:

\[ E_{\text{total}} = \frac{Q}{4 \pi \epsilon_0 R^3} r + \frac{1}{4 \pi \epsilon_0} \frac{\frac{Q}{16}}{r^2} \]

### Step 3: Electric Field Outside the Sphere
For distances greater than \( R \) (i.e., \( r > R \)), the total charge affecting the electric field is the sum of the charge of the sphere and the point charge:

\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q + \frac{Q}{16}}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{\frac{17Q}{16}}{r^2} \]

### Step 4: Analyzing the Graph
1. **For \( r < R \)**: The electric field \( E \) increases linearly with \( r \) due to the uniform charge distribution and the point charge.
2. **At \( r = R \)**: The electric field reaches a maximum value.
3. **For \( r > R \)**: The electric field decreases with the square of the distance \( r^2 \).

### Step 5: Conclusion
Based on the analysis, the graph representing the electric field \( E \) will show:
- A linear increase in \( E \) for \( r < R \).
- A maximum at \( r = R \).
- A decrease as \( r \) increases beyond \( R \).

The correct option is the one that shows this behavior, typically resembling a curve that increases linearly, peaks at \( R \), and then decreases as \( r \) increases.