Question

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x=d and x=−d. The y- and z-dimensions of the slab are very large compared to d, so treat them as infinite. The slab has charge density given by rho(x)=rho0(x/d)2, where rho0 is a positive constant.

Required:
a. Explain why the electric field due to the slab is zero at the center of the slab (x=0).
b. Using Gauss's law, find the magnitude of the electric field due to the slab at the points 0<= x <= d. Express your answer in terms of the variables p, x, d, and e_0.
d. Using Gauss's law, find the magnitude of the electric field due to the slab at the points x>= d. Express your answer in terms of the variables p, x, d, and e_0.
e. What is the direction of the electric field due to the slab at the pints x >= d.

Answer 1

a. The electric field is zero at the center of the slab due to symmetry. b. The magnitude of the electric field at 0<= x <= d is E = (rho0 * d / (2 * e_0)) * (1 - (x^2/d^2)). d. The magnitude of the electric field at x >= d is E = (rho0 * x / (2 * e_0 * d)). e. The direction of the electric field at x >= d is in the positive x-direction.

Step-by-step explanation:

a. The electric field is zero at the center of the slab because it is symmetric. The electric field contributions from each side of the slab cancel each other out.

b. To find the electric field at the points 0<= x <= d, we can use Gauss's law. By choosing a Gaussian surface in the shape of a cylinder with one end inside the slab and the other end outside the slab, we can determine the flux through the Gaussian surface. The flux is equal to the charge enclosed divided by the permittivity of free space multiplied by the area of the Gaussian surface. Solving for the electric field, we find that the magnitude of the electric field is E = (rho0 * d / (2 * e_0)) * (1 - (x^2/d^2)).

d. To find the electric field at the points x >= d, we again use Gauss's law with a Gaussian surface in the shape of a cylinder. This time, the entire slab is enclosed by the Gaussian surface. The charge enclosed is equal to the total charge of the slab, which can be found by integrating the charge density over the volume of the slab. Solving for the electric field, we find that the magnitude of the electric field is E = (rho0 * x / (2 * e_0 * d)).

e. The direction of the electric field at the points x >= d is in the positive x-direction.