Flux of an Electric field with a constant charge density

In summary: Ok, so I understand that the equation for electric flux is something like this, but I still don't know how to find the normal vector. In summary, the problem is that the Electric field and the normal vector are both pointing away from the z axis, so the angle between them is zero. You can solve for the normal vector by solving for the angle between the electric field and the surface normal. Once you have the normal vector, you can use it to find the flux of electric field through the cylinder.
  • #1
want2learn!
5
0
This is actually a problem in my Vector Calc class, but is very fitting for this section.

I have been trying to understand this problem, and cannot seem to figure out where to go.

The z-axis carries a constant electric charge density of λ units of charge per unit length with λ > 0. The resulting electric field is given by

E=5λ*xi+yj/x^2+y^2

Compute the flux of outward through the cylinder x^2 + y^2 ≤ R^2, for 0 ≤ z ≤ h. Use lambda for λ.


I have been having pretty good success with most of the problems from this section, but I cannot seem to set this guy up.

Since both the normal vector, and the electric field are pointing radially away from the z axis, the angle between them is 0.

Can someone give me a step in the right direction please? I have a few problems that are driving me nuts.
 
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  • #2
Hi want2learn, welcome to PF!:smile:

Since you are asked to calculate the electric flux, a good place to start might be to write down the definition (in equation form) of electric flux...don't you think?:wink:
 
  • #3
Ok, when I said I don't know where to start, I did not mean that basic.

I know Gauss's electric flux equation. I know that it involves the Electric field, the normal vector to the area, and the area of the cylinder (2*pi*r*h).

How can I find the normal vector in this problem?
 
  • #4
Draw a picture...the cylinder has 3 surfaces, what are they? How does one normally find the unit normal to a surface?...Do that.
 
  • #5
I drew a picture, and have everything set up. the Surface Area of the cylinder would be 2pi*r^2+2pi*r*h. Since the charge is located along the z axis, and the cylinder is oriented around the z axis, the top and bottom of the cylinder should not be used in the SA. So, the surface area would just be 2*pi*r*h, right?

And, normally I would cross two vectors to find the normal vector, and I am not getting how you find it on a cylinder like this. Since the angle between the normal and the electric field is 0, does it end up just being the Electric field*Surface Area ?
 
  • #6
How do you know that the angle between the electric field and the surface normal is zero if you can't find the surface normal?

Also, the equation for flux involves an integral doesn't it?...What makes you think the result of that integral will just be electric field*surface area?
 

FAQ: Flux of an Electric field with a constant charge density

What is the definition of "flux" in the context of an electric field with a constant charge density?

Flux is a measure of the flow of electric field lines through a given surface. In the context of an electric field with a constant charge density, it represents the amount of electric field passing through a unit area of the surface per unit time. It is typically denoted by the symbol Φ and has units of volts per meter squared (V/m2).

How is the flux of an electric field with a constant charge density calculated?

The flux of an electric field with a constant charge density can be calculated using the formula Φ = E * A, where E is the magnitude of the electric field and A is the area of the surface perpendicular to the direction of the electric field. This formula assumes that the electric field is constant over the entire surface.

What is the relationship between the flux of an electric field with a constant charge density and the charge enclosed within the surface?

According to Gauss's Law, the flux of an electric field with a constant charge density is directly proportional to the net charge enclosed within the surface. This means that as the charge enclosed increases, the flux will also increase. In other words, the flux is a measure of the strength of the electric field due to the enclosed charge.

Can the flux of an electric field with a constant charge density be negative?

Yes, the flux of an electric field with a constant charge density can be negative. This occurs when the direction of the electric field is opposite to the normal vector of the surface. In other words, the electric field lines are entering the surface rather than exiting it. This can happen, for example, near a negative charge enclosed within the surface.

What are some practical applications of understanding the flux of an electric field with a constant charge density?

Understanding the flux of an electric field with a constant charge density is essential for many practical applications, such as designing electrical circuits, analyzing the behavior of electric fields in various materials, and predicting the behavior of charged particles in electric fields. It is also crucial in understanding the behavior of lightning and other atmospheric electricity phenomena.

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