Find the electric flux through the parabolodial surface

In summary, there are two types of electric flux problems discussed in the textbook. The first problem involves finding the electric flux through a paraboloidal surface due to a uniform electric field. The second problem involves an infinitely long line charge and constructing a sphere with a radius less than the distance from the charge. The flux through the surface of the sphere is found to be zero. There is some discussion about whether the flux through the paraboloidal region is also zero, but it is determined that the net flux is indeed zero while the flux through the paraboloidal region equals the flux through the circular region.
  • #1
oneplusone
127
2
In my textbook there are two types of electric flux problems:

1. Find the electric flux through the parabolodial surface due to a uniform electric field of magnitude E_0 in the direction as shown.

Here they use ##\pi r^2 * E = \Phi ##


2. An infinitely long line charge having a uniform charge per unit length ##\laambda ## lies a distance d from point O. construct a sphere with center O, and radius less than d. Find the flux thru the surface of the sphere.

Here the answer is 0 since q enclosed is 0.


But for #1, wouldn't it be zero also since the paraboloid has no charge contained inside of it?
 
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  • #2
A paraboloid is not a Gaussian surface, so Gauss' law doesn't apply.
 
  • #3
More explicitly, a paraboloid is an open surface, not a closed one. A Gaussian surface for use in Gauss's Law must be a closed surface.
 
  • #4
I think he/she is talking about a closed paraboloid such as this one.

pse6.24.p13.png


They may be talking about the single surface of the paraboloid, not the net total of all the surfaces. Since the flux coming into the cap is negative, and the sum of all the flux passing through the surfaces must be zero:

0 = [itex]\Phi[/itex]EndCircle + [itex]\Phi[/itex]Paraboloid = -E0∏r2 + [itex]\Phi[/itex]Paraboloid

Add the E0∏r2 to both sides:

[itex]\Phi[/itex]Paraboloid = E0∏r2

I don't know if you can do this; someone else should verify as I have had no formal education on gauss's first law.

Edited:
oneplusone: We are both in tenth grade (I think). cool.
 
Last edited:
  • #5
As in, perhaps the question was misinterpreted and we are asking for just the flux passing through the paraboloid portion, basically equivalent to the flux passing through the end circle.
 
  • #6
@Above, yes that is the same question! (where'd you find the diagram??)

So am i correct in saying that the NET flux is zero, but the flux thru the parabolodial region equals the flux thru the circular region?
 
  • #7
oneplusone said:
the NET flux is zero, but the flux thru the parabolodial region equals the flux thru the circular region?

As far as I can tell, yes.

I got the image by going onto google images and searching "paraboloid flux".
 

FAQ: Find the electric flux through the parabolodial surface

What is electric flux?

Electric flux is a measure of the electric field passing through a surface. It is defined as the product of the electric field and the area of the surface, perpendicular to the field.

How is electric flux calculated?

The electric flux through a surface is calculated by taking the dot product of the electric field and the area vector of the surface. This can be represented mathematically as Φ = E · A, where Φ is the electric flux, E is the electric field, and A is the area vector.

What is the significance of finding electric flux through a paraboloidal surface?

Finding the electric flux through a paraboloidal surface can help us understand the strength and direction of the electric field in a given region. It can also be useful in determining the total charge enclosed by the surface.

What are the units of electric flux?

The SI unit of electric flux is volt-meters (V·m), while the CGS unit is statvolt-centimeters (statV·cm). Both units are equivalent to coulombs (C).

Can electric flux be negative?

Yes, electric flux can be negative. This occurs when the electric field and area vector are in opposite directions, resulting in a negative dot product. A negative electric flux represents an inward flow of electric field, while a positive electric flux represents an outward flow.

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