Respuesta :
Answer:
Electric field at (x, y, z): [tex]E=\frac{\rho x}{\epsilon_o}[/tex]
Explanation:
(a) The slab is of an insulating material and has a uniform charge distribution. We can visualize this as infinite number of point charges, distributed throughout the slab, equally spaced apart. So if we (hypothetically) start to calculate the electric field due to each charge at x = 0, we shall always find a charge at a mirrored position about the x = 0 plane (within x = -d and x = d) and hence will cancel out the electric field.
A simpler example would be an infinitely long wire of uniform charge distribution. Any point on the wire will have zero electric field has there are essentially equal number of charges on either side (the length of the wire being infinitely long)
(b) Let us take a cylinder as a Gaussian surface with base area A. We shall take advantage of the symmetry about x = 0 and shall position the cylinder perpendicular to the y-z plane with x = 0 being the mid-point. Now the electric flux will only flow out through the 2 bases of the cylinder. This is because the slab has infinite dimensions along y and z-axes (think of an infinite sheet of charge) and the electric field always starts out perpendicular to any surface of charges. If the Electric field at some point on the base of the cylinder be E, then total outgoing flux = 2EA
[tex]\rho[/tex] is the charge density, hence, [tex]Q_{enclosed}=\rho\times volume=2\rho Al[/tex]
where [tex]2l[/tex] is the length of the cylinder and [tex]l[/tex] is the x-coordinate.
Therefore, using Gauss's law,
[tex]2EA=\frac{2\rho Al}{\epsilon_o}[/tex]
or, [tex]E=\frac{\rho l}{\epsilon_o}[/tex]
or, [tex]E=\frac{\rho x}{\epsilon_o}[/tex]
where, [tex]\epsilon_o[/tex] = permittivity of free space.
