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 represent quanlitatively, the variation of electric field intensity E with distance r from the centre.

Options

A.
B.
C.
D.

Explanation

To solve the problem of the electric field intensity \( E \) variation 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 System
We have a spherical insulator of radius \( R \) uniformly charged with charge \( Q \) and a point charge \( \frac{Q}{16} \) at its center. The electric field \( E \) will vary based on the distance \( r \) from the center.

### Step 2: Analyze the Electric Field Inside the Sphere
For a uniformly charged sphere, the electric field inside the sphere (at a distance \( r < R \)) can be calculated using Gauss's law. The electric field \( E \) at a distance \( r \) from the center is given by:

\[ E = \frac{1}{4\pi \epsilon_0} \left( \frac{Q_{\text{enc}}}{r^2} \right) \]

Where \( Q_{\text{enc}} \) is the charge enclosed within the Gaussian surface of radius \( r \). The charge enclosed is the charge due to the point charge at the center and the charge distributed in the volume of the sphere.

### Step 3: Calculate the Enclosed Charge
The charge enclosed \( Q_{\text{enc}} \) at a distance \( r \) is given by:

\[ Q_{\text{enc}} = \frac{Q}{V} \cdot V_{\text{enc}} + \frac{Q}{16} \]

Where \( V \) is the total volume of the sphere and \( V_{\text{enc}} \) is the volume of the sphere of radius \( r \):

\[ V = \frac{4}{3} \pi R^3, \quad V_{\text{enc}} = \frac{4}{3} \pi r^3 \]

Thus,

\[ Q_{\text{enc}} = \left( \frac{Q}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi r^3 \right) + \frac{Q}{16} = Q \frac{r^3}{R^3} + \frac{Q}{16} \]

### Step 4: Electric Field Inside the Sphere
Substituting \( Q_{\text{enc}} \) back into the electric field equation gives:

\[ E = \frac{1}{4\pi \epsilon_0} \left( \frac{Q \frac{r^3}{R^3} + \frac{Q}{16}}{r^2} \right) \]

This shows that the electric field inside the sphere will increase with \( r \) until it reaches the surface.

### Step 5: Electric Field Outside the Sphere
For \( r > R \), the entire charge \( Q + \frac{Q}{16} \) can be treated as a point charge at the center. Therefore, the electric field outside the sphere is given by:

\[ E = \frac{1}{4\pi \epsilon_0} \left( \frac{Q + \frac{Q}{16}}{r^2} \right) = \frac{1}{4\pi \epsilon_0} \left( \frac{17Q/16}{r^2} \right) \]

### Step 6: Graphical Representation
1. **Inside the Sphere (0 < r < R)**: The electric field starts from zero, increases as \( r \) increases, and reaches a maximum at \( r = R \).
2. **At the Surface (r = R)**: The electric field will have a maximum value.
3. **Outside the Sphere (r > R)**: The electric field decreases as \( \frac{1}{r^2} \).

### Conclusion
The graph that best represents the variation of electric field intensity \( E \) with distance \( r \) from the center will show an increase in \( E \) from \( r = 0 \) to \( r = R \), followed by a decrease for \( r > R \).