Nonuniform volume charge density

[shinobi12]

Homework Statement

A slab of insulating material has a nonuniform volume charge density
given by rho = Cx2, where C is a positive constant and x is measured from the slab’s center. The slab is infi…nite in the y and z directions.

(a) Find the electric fi…eld for |x| > d=2, that is, in the regions exterior to the slab.
(b) Find the electric …field for |x| < d=2, that is, in the interior region of the slab.

Homework Equations

Not sure

The Attempt at a Solution

I tried thinking in terms of gauss' law but I'm term how to tackle this problem

 

[gabbagabbahey]

Gauss' Law sounds good...what kind of symmetry does this problem possess? What type of Gaussian surface do you use when that type of symmetry is present?

 

[thomate1]

.

I would approach this problem by first understanding the concept of volume charge density. Volume charge density (ρ) is defined as the amount of charge (Q) per unit volume (V) of a given material, expressed mathematically as ρ = Q/V. In this case, the volume charge density is nonuniform, meaning it varies with position in the slab.

To solve for the electric field in the regions exterior to the slab (|x| > d/2), we can use Gauss' Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε0). In this case, we can consider a Gaussian surface in the shape of a cylinder with the slab's center as the axis. The electric field will be constant on the surface of the cylinder, and we can use the formula for the electric flux to solve for the electric field outside the slab.

To solve for the electric field in the interior region of the slab (|x| < d/2), we can use the same approach, but the Gaussian surface will now be a sphere with the slab's center as the center of the sphere. Again, the electric field will be constant on the surface of the sphere, and we can use the electric flux formula to solve for the electric field inside the slab.

In both cases, we will need to integrate the volume charge density function (ρ = Cx^2) over the appropriate volume to find the total enclosed charge. From there, we can use the electric flux formula to solve for the electric field.

In summary, to solve this problem as a scientist, I would use the concept of volume charge density and Gauss' Law to solve for the electric field in both the exterior and interior regions of the slab.