# Flux through a Cone

1. Apr 15, 2007

### kaotak

Source: Physics and Scientists for Engineers, Ch. 24 #7

A cone of radius R and height h sits on a horizontal table. A uniform electric field parallel to the table penetrates the surface of the cone. What is the flux entering the cone?

Diagram:
(N.B. the dots in the cone are just to give it shape)

E
-->.../.\
-->../...\
-->./.....\ <---- cone
-->/____\___________ <- table

The back of the book gives an answer of flux_entering=EhR, which I don't get. I don't see why there isn't a pi in that answer, since the surface area of the face of the cone that's being penetrated certainly does have a pi in it.

My attempt at a solution yields flux_entering=pi*E*h*r/3. Hah, it's close... but not.

EDIT: The field is uniform, I forgot to mention that.

Last edited by a moderator: Apr 15, 2007
2. Apr 15, 2007

### Mathgician

what is the integral equation and what does it equal? Need to see the steps of your attempt.

3. Apr 15, 2007

### kaotak

A is the area of the face of the cone that the electric field penetrates, which is half of the surface area of the cone excluding the bottom area. So

Last edited by a moderator: Apr 15, 2007
4. Apr 15, 2007

### hage567

I'm pretty sure that is not the formula for the surface area of a cone. I'd double check it, something doesn't look right about it.

5. Apr 15, 2007

### kaotak

You're right. It's SA=pi*R*S (neglecting the base) where S is the slant height. But that still doesn't explain why there's no pi in the textbook solution.

6. Apr 15, 2007

### Päällikkö

You need to consider the perpendicular part as you're dealing with a dot product. So the surface area you should calculate is not that of half a cone but... what?

That was a bit confusing. Anyways
$$\int_S \mathbf{E}\cdot d\mathbf{A} = E\int_S dA$$
is wrong.

Last edited: Apr 15, 2007
7. Apr 15, 2007

### hage567

OK actually you don't need the surface area of cone. You have to take the projection of the cone onto a plane normal to the electric field. That's the area you are interested in.

8. Apr 15, 2007

### kaotak

That's what I thought too but... by that logic can't you say that the shape of a surface doesn't matter when considering the flux?

9. Apr 15, 2007

### bdrosd

I am probably missing something but isn't the flux going to be zero since you have the electric field entering on one side and exiting the cone on the other? The normal direction would be pointing outwards and thus the sign of the dot product would reverse from one side to the other?

10. Apr 15, 2007

### Päällikkö

That is the point of flux really, and the mathematical formulation to achieve this is the dot product. Think of water flow, it really doesn't matter what shape you stick in there, the same amount of water will flow through. If you throw in a closed surface, the same amount of water that flows in, flows out, unless there happens to be a source or a drain inside the surface. In electromagnetism this is called Gauss's law, which you've probably encountered, or soon will.

Only the part going in was asked in the problem.

11. Apr 15, 2007

### kaotak

Oh, I see, thanks. I thought that was only true for a gaussian surface enclosing a charge, but it's also true for the effects of an external charge on the flux entering a gaussian surface (whoo, that was a mouthful).

12. Jul 27, 2008

### sstewari

Yes, you require to take only the projection of area of cone on a plane normal to electric field and it is a triangle for the cone. Area of the triangle = 1/2 * 2r * h = rh
Hence flux = Erh.

13. Jun 4, 2010

### stronghold.mr

i am still not able to get it.................

I am messed up with this concept of projection of area of different surfaces and flux through different surfaces................

Please someone explain it with some examples such as -

1. Hemisphere
2. Frustum
3. Cone

we have to find flux through all these surfaces