A thin-walled spherical conducting shell S of radius R is given charge Q. The same amount of charge is also placed at its center C. Which of the following statements are correct?

Options

A. On the outer surface of S, the charge density is `Q//2piR^2`.
B. The electric field is zero at all points inside S.
C. At a point just outside S, the electric field is double the field at a point just inside S in the cavity.
D. At any point inside S (i.e., within its cavity), the electric field is inversely proportional to the square of its distance from C.

Explanation

To solve the problem, we need to analyze the situation involving a thin-walled spherical conducting shell with a charge Q on its surface and an equal charge Q at its center. We will evaluate each statement provided in the question.

### Step-by-Step Solution:

1. **Understanding Charge Distribution**:
- The conducting shell has a total charge Q on its outer surface.
- When a charge Q is placed at the center, it induces a charge of -Q on the inner surface of the shell (due to electrostatic shielding).
- Consequently, the outer surface of the shell will have a charge of +Q (from the original charge) plus the induced -Q from the inner surface, resulting in a total charge of +Q on the outer surface.

2. **Electric Field Inside the Shell**:
- Inside a conductor in electrostatic equilibrium, the electric field is zero. However, since there is a charge at the center, the electric field inside the cavity (the space between the center charge and the inner surface of the shell) is not zero.
- The electric field at a point inside the cavity (due to the charge at the center) can be calculated using Gauss's Law. The electric field \( E \) at a distance \( r \) from the center is given by:
\[ E = \frac{kQ}{r^2} \]
- Here, \( k \) is Coulomb's constant.

3. **Electric Field Just Outside the Shell**:
- Just outside the shell, the electric field is influenced by both the charge at the center and the charge on the outer surface.
- The electric field just outside the shell (at distance \( R \)) can be calculated as:
\[ E_{\text{outside}} = \frac{kQ}{R^2} + \frac{kQ}{R^2} = \frac{2kQ}{R^2} \]
- This shows that the electric field just outside the shell is double that of the electric field just inside the shell.

4. **Electric Field Inside the Cavity**:
- The electric field inside the cavity is inversely proportional to the square of the distance from the charge at the center, as shown by the formula \( E = \frac{kQ}{r^2} \).

### Conclusion:
Based on the analysis, we can evaluate the correctness of the statements:

- **Statement A**: Correct. The surface charge density on the outer surface is \( \sigma = \frac{Q}{4\pi R^2} \).
- **Statement B**: Incorrect. The electric field inside the cavity is not zero; it is \( \frac{kQ}{r^2} \).
- **Statement C**: Correct. The electric field just outside the shell is indeed double that just inside the shell.
- **Statement D**: Correct. The electric field inside the cavity is inversely proportional to the square of the distance from the center charge.

Thus, the correct statements are A, C, and D.