Answer:
please read the answer below
Explanation:
To find the electric field you can consider the Gaussian law for a cylindrical surface inside the slab.
[tex]\int E dA=EA_{G}=\frac{Q_{int}}{\epsilon_o}[/tex]
[tex]Q_{int}=\rho V_{G}[/tex]
where Qint is the charge inside the Gaussian surface, AG is the area of the surface and rho is the charge density of the slab.
By using the formula for the volume of a cylinder you obtain:
[tex]V_{G}=\pi r^2h[/tex]
where h is the height. If you assume that the slab is in the interval (-zo<z<z0) you can write VG:
[tex]V_{G}=\pi r^2 z[/tex]
Finally, by replacing in the expression for E you get:
[tex]E=\frac{Q_{int}}{\epsilon_o A_G}=\frac{Q_{int}}{\epsilon_o \pi r^2}\frac{z}{z}=\frac{\rho z}{\epsilon_o}[/tex]
[tex]E=\rho z/\epsilon_o[/tex]
hence, for z>0 you obtain E=pz/eo > 0
for z<0 -> E=pz/eo < 0
