Diagram
Deducing electric field expression
(i) To the left of first sheet
(ii) To the right of second sheet
(iii) Between the two sheets
Diagram :

Electric field in the region left of first sheet
${\mathit{E}}_{\mathrm{I}}={\mathit{E}}_1+{\mathit{E}}_2$
${\mathit{E}}_{\mathrm{I}}=\frac{\mathit{\sigma }}{{\mathit{\epsilon }}_0}-\frac{\mathit{\sigma }}{2{\mathit{\epsilon }}_0}$
${\mathit{E}}_{\mathrm{I}}=+\frac{\mathit{\sigma }}{2{\mathit{\epsilon }}_0}$
It is towards right,
Electric field in the region to the right of second sheet
${\mathit{E}}_{\mathrm{II}}=\frac{\mathit{\sigma }}{2{\mathit{\epsilon }}_0}-\frac{\mathit{\sigma }}{{\mathit{\epsilon }}_0}$
${\mathit{E}}_{\mathrm{II}}=-\frac{\mathit{\sigma }}{2{\mathit{\epsilon }}_0}$
It is towards left,
Electric field between the two sheets



Electric field is towards the right.
OR
Diagram
Finding the surface charge density in the inner and outer surface of the shell
Electric field in the cavity
Diagram:

The surface charge density on inner surface of the shell is ${\mathit{\sigma }}_1=-\frac{\mathit{q}}{4\mathit{\pi }{\mathit{r}}_1^2}$
The surface charge density on outer shell is
${\mathit{\sigma }}_2=-\frac{\mathit{Q}+\mathit{q}}{4\mathit{\pi }{\mathit{r}}_2{}^2}$
(b) Consider a Gaussian surface inside the shell, net flux is zero since . According to Gauss's law it is independent of shape and size of shell.