(a) Charge placed at the centre of the hollow sphere is $-q$. Hence, a charge of magnitude $+q$ will be induced to the inner surface. Therefore, total charge on the inner surface of the shell is $+q$. Surface charge density at the inner surface $\sigma_i = \; \frac{\; \text{ Total charge } \;}{\; \text{ Inner surface area } \;} = \; \frac{+q}{4 \pi r_1^2}$ A charge of $-q$ is induced on the outer surface of the sphere. A charge of magnitude $Q$ is placed on the outer surface of the sphere. Therefore, total charge on the outer surface of the sphere is $Q-q$. Surface charge density at the outer surface $\sigma_{\;\text{outer}\;} = \; \frac{\; \text{ Total charge } \;}{\; \text{ Outer surface area } \;} = \; \frac{Q-q}{4 \pi r_2^2}$ (b) Electric field at point lying outside the sphere at a distance $r$ from the centre of the sphere: Applying Gauss theorem $\; \text{ Flux } \; = \phi = \; \frac{\; \text{ Charge enclosed } \;}{\epsilon_o}$ $\; \text{ Or, } \;$ $E \times 4 \pi r^2 \; = \; \frac{Q-q}{\epsilon_n}$ $\therefore E \; = \; \frac{Q-q}{4 \pi r^2 \epsilon_0}$ OR (a) Electric field due to an infinitely long straight wire having uniform linear charge density $\lambda$ : $x=$ distance of the point $P$ from the wire where the electric field is to be evaluated $E=$ electric field at the point $P$ A Gaussian cylinder of length $l$, radius $x$ is considered. An infinitesimally small area $d s$ on the Gaussian surface is considered. Electric field is same at all points on the curved surface of the cylinder and directed radially outward. So, $E$ and $d s$ are along the same direction. The total electric flux $(\phi)$ through curved surface $= \int E d s \cos \theta$ Since $E$ and $d s$ are along the same direction, so $\theta=$ $0^{\circ}$ So, $\; \; \phi = E(2 \pi x l)$ The net charge enclosed by Gaussian surface is, $q=$ $\lambda l$ $\therefore$ By Gauss's law, $\phi = \; \frac{1}{\epsilon_0} q$Or, $\therefore \; \frac{1}{\epsilon_0} q = E(2 x r l)$ $\therefore E = \; \frac{\lambda}{2 \pi x \epsilon_0}$ (b)