Find the electric flux through the parabolodial surface

[oneplusone]

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?

 

[Dale]

A paraboloid is not a Gaussian surface, so Gauss' law doesn't apply.

 

[jtbell]

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.

 

[TheDemx27]

I think he/she is talking about a closed paraboloid such as this one.

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 = $\Phi$_EndCircle + $\Phi$_Paraboloid = -E₀∏r² + $\Phi$_Paraboloid

Add the E₀∏r² to both sides:

$\Phi$_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.

 

[TheDemx27]

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.

 

[oneplusone]

@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?

 

[TheDemx27]

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".