(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

[tex]\phi_E = EA[/tex]

[tex]\phi_E = EA' = EA \cos \theta[/tex]

3. 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 iswhy?Here is the solution from the cited page:

A = .5 * b * h, b = 2R

so [tex]\phi_E = E \frac{1}{2} 2 R h = E R h[/tex]

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?

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# Electric flux on half a conic

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