Given setup

• A conducting spherical shell has inner radius $a$ and outer radius $b$.
• Net charge on the conductor is $Q$.
• A dipole of moment $\vec P$ is placed at the centre of the cavity.

We must determine which statement is correct.

Key electrostatic facts for a conductor

In electrostatic equilibrium:

1. Electric field inside the conducting material is zero.
2. The conductor is an equipotential.
3. Charges may redistribute on inner and outer surfaces.

Charge induced on the inner surface

The dipole consists of charges $+q$ and $-q$ separated by a small distance, so the net charge inside the cavity is zero.

Using Gauss's law for a Gaussian surface lying within the conductor just outside the inner cavity: $\oint \vec E\cdot d\vec A = 0 \implies q_{\text{enclosed}}=0$

Since the dipole has net enclosed charge zero, the total induced charge on the inner surface is zero.

But zero total induced charge does not mean zero charge density everywhere. Because the dipole creates a non-uniform field in the cavity, the inner surface acquires a non-uniform induced distribution with positive charge on one side and negative on the other, summing to zero.

Therefore:

• A is false: inner surface charge density is not uniform, and certainly not $\dfrac{Q/2}{4\pi a^2}$.
• B is false: charge density on the inner surface is not zero everywhere.

Charge on the outer surface

The conductor has total charge $Q$. Since total charge on inner surface is $0$, the total charge on the outer surface must be $Q_{\text{outer}}=Q.$

Does outer surface distribution depend on $|\vec P|$?

The conductor is spherical, and the outer surface is a spherical equipotential. Outside the conductor, there is no charge except the total charge residing effectively on the outer surface.

Since the outer boundary is a sphere at constant potential, by uniqueness theorem the potential outside must be spherically symmetric: $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}, \qquad r\ge b$

Hence the external field is $\vec E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat r.$

So the outside field is completely independent of the dipole moment $\vec P$.

Therefore:

• C is false: outer surface distribution / external field does not depend on $|\vec P|$.

Field outside the shell

From the expression above, the field outside is exactly the same as that due to a point charge $Q$ placed at the centre.

Therefore:

• D is true.

Final conclusion

The only correct option is: $\boxed{D}$

Comparison with stored answer

Stored correct answer: $D$

My derived answer also is $D$, so they agree.