The problem involves a solid sphere with charge $-4Q$ inside a spherical conducting shell with charge $+8Q$, and asks for the electric field outside the shell.

Key points to consider:

• Gauss’s Law states that the electric flux through a closed surface is proportional to the enclosed charge: $\Phi_E = \frac{Q_\text{enc}}{\epsilon_0}$ where $Q_\text{enc}$ is the charge enclosed by the surface, and $\epsilon_0$ is the permittivity of free space.

• The electric field outside a spherical shell behaves as though all the charge were concentrated at the center of the shell. For a spherical shell with charge $+8Q$, the field outside the shell will be equivalent to the field due to a point charge of $+8Q$ at the center.

Calculation:

In the region outside the spherical shell, the total enclosed charge is the charge of the shell itself, which is $+8Q$. Applying Gauss's law, the electric field at a distance $r$ from the center of the shell is given by: $E = \frac{Q_\text{enc}}{4\pi r^2 \epsilon_0}$ Here, $Q_\text{enc} = +8Q$.

Thus, the electric field outside the shell is: $E = \frac{8Q}{4\pi r^2 \epsilon_0}$ This matches option d: $\frac{4Q}{\pi r^2 \epsilon_0}$

Answer: d. $\frac{4Q}{\pi r^2 \epsilon_0}$

Would you like a further explanation or clarification on any of the steps? Here are some related questions you may find interesting:

1. How does the electric field behave inside the spherical conducting shell?
2. What happens to the electric field inside the solid sphere?
3. How would the electric field change if the charge on the shell were negative instead of positive?
4. What is the potential difference between the center of the sphere and a point outside the shell?
5. Can we apply Gauss’s Law to a non-spherical object, and how would it differ?

Tip: For spherical symmetry, remember that the electric field depends only on the total enclosed charge and the radial distance from the center.