How Is Electric Flux Calculated for a Cone in a Uniform Field?

[lizzyb]

Homework Statement

24.9 A cone of base radius R and height h is located on a horizontal table, and a horizontal uniform electric field E penetrates the cone, as shown here. Determine the electric flux entering the cone.

You can actually view the question and solution at:
http://www.ux1.eiu.edu/~cfadd/1360/24Gauss/HmwkSol.html

Homework Equations

$$\phi_E = EA$$
$$\phi_E = EA' = EA \cos \theta$$

The Attempt at a Solution

My first reaction was to set up a double integeral involving the surface area and cosine but that didn't go very far and then I found the solution on the web. My question is why? Here is the solution from the cited page:

A = .5 * b * h, b = 2R
so $$\phi_E = E \frac{1}{2} 2 R h = E R h$$

It is as if they unrolled half the surface of the cone and formed a triangle out of it which gave the area. Is that correct?

 

[Gokul43201]

No, that's not what they did. They projected the cone onto a plane normal to the E-field, and found the area of this projection.

$E \cdot dA = EAcos \theta$; the dot product is the projection of the area vector along the E-field times the strength of the E-field (recall the construction of the dot/scalar product). So, if you take each little area element (remember that the area vector points normally to the plane of the element), find its projection along the E-direction and then "add up" (integrate) these projections you find yourself with nothing but the projection of the entire surface area (which is nothing but the area of the "shadow" cast by this E-field).

 

[m2003]

Yes, that is correct. The solution on the cited page is using a simplified method to calculate the electric flux. The cone can be thought of as a flat surface that has been curved into a cone shape. By unrolling half of the cone, we can calculate the surface area of the curved surface as a triangle, as you mentioned. This makes the calculation much simpler and allows us to use the formula for the flux through a flat surface, \phi_E = EA.

Another way to think about it is that the electric field is perpendicular to the surface of the cone at every point, so we can use the simpler formula for a flat surface. This is because the electric flux is a measure of the number of electric field lines passing through a given area, and in this case, all of the field lines are perpendicular to the surface.

In general, it is important to understand the concept behind the solution and how it relates to the problem at hand, rather than just memorizing formulas. In this case, understanding the relationship between the electric field and the surface of the cone allows us to simplify the calculation and arrive at the correct solution.