1. Problem 5: Given point charges $q_1, q_2, q_3$ inside a spherical Gaussian surface $S$ of radius $R$, and charge $q_4$ outside $S$, we analyze the total electric flux through $S$ using Gauss's law. 2. Gauss's law states: $$\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$$ where $Q_{enc}$ is the total charge enclosed by surface $S$. 3. Since $q_1, q_2, q_3$ are inside $S$ and $q_4$ is outside, $$Q_{enc} = q_1 + q_2 + q_3$$ 4. Therefore, the total flux is: $$\oint_S (\vec{E}_1 + \vec{E}_2 + \vec{E}_3) \cdot d\vec{A} = \frac{q_1 + q_2 + q_3}{\epsilon_0}$$ 5. Charges outside the Gaussian surface ($q_4$) do not contribute to the flux. 6. Hence, the correct option is (a). --- 7. Problem 6: A solid conducting sphere of radius $a$ has net charge $2Q$. A concentric spherical conducting shell with inner radius $b$ and outer radius $c$ carries net charge $-Q$. They are connected by a wire briefly; find surface charge densities on the shell's inner and outer surfaces. 8. When connected by a wire, charges redistribute so that the solid sphere and shell are equipotential. 9. The inner surface of the shell faces the sphere. To cancel the field inside the material of the conductor (shell), it must induce charge $-2Q$ on its inner surface, opposite to sphere charge. 10. Since the shell's total charge is $-Q$, if inner surface has $-2Q$, the outer surface must hold: $$Q_{outer} = -Q - (-2Q) = +Q$$ 11. Surface charge densities are: - Inner surface charge density: $$\sigma_{inner} = \frac{-2Q}{4\pi b^2}$$ - Outer surface charge density: $$\sigma_{outer} = \frac{+Q}{4\pi c^2}$$ 12. These ensure the shell is neutral inside and the field outside corresponds to the net charges. Final answers: - Problem 5: Option (a) is correct. - Problem 6: $$\sigma_{inner} = \frac{-2Q}{4\pi b^2}, \quad \sigma_{outer} = \frac{Q}{4\pi c^2}$$