# Using Gauss Law to Solve a Planar Slab of Charge Problem

In summary, the problem involves a planar slab of charge with a uniform charge density and the goal is to find the electric field vector in the region between -a and a using Gauss's law. The surface of integration chosen should be a Gaussian surface placed in such a way that the electric field at x = 0 is zero due to symmetry. The enclosed charge can be determined by multiplying the charge density by the volume enclosed by the Gaussian surface.

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

We have A planar slab of charge with a charge density ρvvosin(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*Dx=Qenclosed
Then I am stuck.
The Qenclosed integration give me some problem.
Please can someone help me to establish the Gauss law to solve the problem?

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?)

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

## 1. What is Gauss Law and how is it used to solve planar slab of charge problems?

Gauss Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge. It can be used to solve planar slab of charge problems by calculating the electric field at a point outside or inside the slab by considering the symmetry of the problem and using a Gaussian surface.

## 2. What is a planar slab of charge?

A planar slab of charge is a two-dimensional object with a uniform distribution of electric charge on its surface. It is often represented as a thin sheet with a finite thickness and can have a positive, negative, or neutral charge.

## 3. What is a Gaussian surface and how is it used in Gauss Law?

A Gaussian surface is an imaginary surface that is used to simplify the calculation of electric fields using Gauss Law. It is a closed surface that encloses the charge distribution and has a known electric field value at every point on its surface. By choosing a Gaussian surface with the appropriate symmetry, the calculation of the electric field can be simplified to a single variable integration.

## 4. Can Gauss Law be used to solve problems with non-uniform charge distributions?

Yes, Gauss Law can be used to solve problems with non-uniform charge distributions. However, in these cases, a more complex Gaussian surface may need to be chosen in order to accurately describe the electric field at a specific point.

## 5. Are there any limitations or assumptions when using Gauss Law to solve planar slab of charge problems?

One limitation of using Gauss Law to solve planar slab of charge problems is that it only applies to problems with static electric fields. Additionally, it assumes that the charge distribution is continuous and there are no external electric fields present. It also assumes that the material of the slab is a perfect insulator and has no conductivity.

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