Using Gauss Law to Solve a Planar Slab of Charge Problem

[brad sue]

Hey, I need help to use the Gauss law in this problem:

We have A planar slab of charge with a charge density ρ_v=ρ_vosin(2*pi/(2*a)),for -a<x<a
the thickness of the stab is 2*a.
the horizontal y-axis passes through the middle of the stab.
the x-axis is vertical
a) Find the electric field vector in the region -a<x<a.

I know I need to use the Gauss law.
But I have some problems to establish the integral.
What surface should I take?

I tried the cylinder as surface of integration.
I got:
FLUX=2(pi)x*L*D_x=Q_enclosed
Then I am stuck.
The Q_enclosed integration give me some problem.
Please can someone help me to establish the Gauss law to solve the problem?

 

[Saketh]

Tactical use of Gaussian surfaces is a skill one acquires after suffering through many Gauss's law problems. So don't fret if you don't get it immediately.

In this case, since the planar charge distribution has uniform charge density (ρv=ρvosin(2*pi/(2*a)) is a constant), the electric field at x = 0 would be zero, because of symmetry. Convince yourself that this is true. Based on this, figure out where to put your Gaussian surface.

When you say I tried "the cylinder", success often depends on where you put its faces.

The enclosed charge is simple enough to determine -- just multiply the charge density by the volume enclosed by the Gaussian surface. (Or did you mean to give a charge distribution that depends on x?)

 

[m2003]

First of all, it is great that you are attempting to use Gauss's law to solve this problem. Gauss's law is a powerful tool in solving problems related to electric fields and charges.

In this problem, we have a planar slab of charge with a non-uniform charge density. To use Gauss's law, we need to choose a closed surface in which the electric flux can be easily calculated. In this case, a cylinder is a good choice for the surface of integration. However, the dimensions of the cylinder need to be carefully chosen.

Since the slab is infinite in the y-direction, we can choose a cylinder with a small radius r and a length L, such that the cylinder encloses the slab completely. The cylinder should also be symmetric about the x-axis, as the problem is also symmetric about the x-axis. This will help in simplifying the calculations.

Now, according to Gauss's law, the electric flux through a closed surface is equal to the net charge enclosed by that surface. In this case, the net charge enclosed will be the charge density multiplied by the volume of the slab enclosed by the cylinder. This can be written as:

Flux = Qenclosed = ρv * (2a * 2r * L)

Now, we need to calculate the electric field at any point on the surface of the cylinder. This can be done by using the formula for electric field due to a point charge. We consider a small element of charge dQ on the surface of the cylinder, which will contribute to the electric field at the point. This can be written as:

dE = k * (dQ/r^2)

where k is the Coulomb's constant and r is the distance between the element of charge and the point on the surface. Now, we can integrate this expression over the entire surface of the cylinder to get the total electric field at the point. This can be written as:

E = k * ∫ (dQ/r^2)

Substituting the value of dQ from the expression for the charge density given in the problem, we get:

E = k * ∫ (ρv * (2a * 2r * L)/r^2)

Solving this integral, we get:

E = k * (ρv * 4aL) * ∫ (1/r^2) dr

E = k * (ρv * 4aL) * (-1/r