(a) Charge placed at the centre of the hollow sphere is $-\mathit{q}$. Hence, a charge of magnitude $+\mathit{q}$ will be induced to the inner surface. Therefore, total charge on the inner surface of the shell is $+\mathit{q}$. Surface charge density at the inner surface

A charge of $-\mathit{q}$ is induced on the outer surface of the sphere. A charge of magnitude $\mathit{Q}$ is placed on the outer surface of the sphere. Therefore, total charge on the outer surface of the sphere is $\mathit{Q}-\mathit{q}$. Surface charge density at the outer surface

(b) Electric field at point lying outside the sphere at a distance $\mathit{r}$ from the centre of the sphere:
Applying Gauss theorem

$\mathrm{E}\times 4\mathit{\pi }{\mathrm{r}}^2=\frac{\mathit{Q}-\mathit{q}}{{\mathit{\epsilon }}_{\mathrm{n}}}$
$\therefore \mathrm{E}=\frac{\mathit{Q}-\mathit{q}}{4\mathit{\pi }{\mathrm{r}}^2{\mathit{\epsilon }}_0}$
OR
(a) Electric field due to an infinitely long straight wire having uniform linear charge density $\mathit{\lambda }$ : $\mathit{x}=$ distance of the point $\mathrm{P}$ from the wire where the electric field is to be evaluated
$\mathit{E}=$ electric field at the point $\mathit{P}$
A Gaussian cylinder of length $\mathit{l}$, radius $\mathit{x}$ is considered.
An infinitesimally small area $\mathit{d}\mathit{s}$ 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, $\mathit{E}$ and $\mathit{d}\mathit{s}$ are along the same direction.
The total electric flux $\left(\mathit{\varphi }\right)$ through curved surface $=\int \mathit{E}\mathit{d}\mathit{s}\cos \mathit{\theta }$
Since $\mathit{E}$ and $\mathit{d}\mathit{s}$ are along the same direction, so $\mathit{\theta }=$ $0^{\circ }$
So, $\mathit{\varphi }=\mathit{E}\left(2\mathit{\pi }\mathit{x}\mathit{l}\right)$
The net charge enclosed by Gaussian surface is, $\mathit{q}=$ $\mathit{\lambda }\mathit{l}$
$\therefore$ By Gauss's law,
$\mathit{\varphi }=\frac{1}{{\mathit{\epsilon }}_0}\mathit{q}$
Or, $\therefore \frac{1}{{\mathit{\epsilon }}_0}\mathrm{q}=\mathit{E}\left(2\mathit{x}\mathit{r}\mathit{l}\right)$
$\therefore \mathrm{E}=\frac{\mathit{\lambda }}{2\mathit{\pi }\mathit{x}{\mathit{\epsilon }}_0}$
(b)