A metallic spherical shell of radius R has a charge -Q on it. A point charge +Q is placed at the centre of the shell. Which of the graphs shown below may correctly represent the variation of electric field E with distance r from the centre of shell ?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the electric field \( E \) at different distances \( r \) from the center of a metallic spherical shell with a charge \( -Q \) and a point charge \( +Q \) placed at its center.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a metallic spherical shell of radius \( R \) with a charge of \( -Q \).
- A point charge \( +Q \) is placed at the center of the shell.
- We need to find the variation of the electric field \( E \) with distance \( r \) from the center.

**Hint**: Visualize the setup with a diagram showing the shell and the point charge.

2. **Electric Field Inside the Shell (r < R)**:
- According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.
- For \( r < R \), the only charge enclosed by a Gaussian surface is the point charge \( +Q \).
- The electric field \( E \) at a distance \( r \) from the center can be calculated using the formula:
\[ E = \frac{kQ}{r^2} \]
- Here, \( k \) is Coulomb's constant.

**Hint**: Remember that the electric field inside a conductor is zero, but inside the cavity (where the point charge is), it is determined by the point charge.

3. **Electric Field on the Surface of the Shell (r = R)**:
- At the surface of the shell, the electric field is given by:
\[ E = \frac{kQ}{R^2} \]
- This is the electric field just at the surface.

**Hint**: The electric field is not zero at the surface; it is determined by the charge enclosed.

4. **Electric Field Outside the Shell (r > R)**:
- For \( r > R \), the total charge enclosed by a Gaussian surface is \( +Q + (-Q) = 0 \).
- Therefore, the electric field \( E \) outside the shell is zero:
\[ E = 0 \]

**Hint**: For distances greater than the radius of the shell, consider the net charge enclosed.

5. **Graphing the Electric Field**:
- For \( r < R \), the electric field \( E \) is inversely proportional to \( r^2 \) and approaches infinity as \( r \) approaches zero.
- At \( r = R \), the electric field is at a finite value \( \frac{kQ}{R^2} \).
- For \( r > R \), the electric field drops to zero.

**Hint**: Sketch the graph based on the behavior of \( E \) in the three regions: inside, on the surface, and outside the shell.

### Conclusion:
The graph representing the variation of electric field \( E \) with distance \( r \) from the center of the shell should show:
- An increasing electric field as \( r \) approaches zero,
- A finite value at \( r = R \),
- A drop to zero for \( r > R \).

The correct graph would depict these characteristics, confirming that the electric field is zero outside the shell and behaves according to the inverse square law inside the cavity.