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Electric flux on half a conic

  • Thread starter lizzyb
  • Start date
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1. 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

2. Homework 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 is why? 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?
 

Gokul43201

Staff Emeritus
Science Advisor
Gold Member
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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.

[itex]E \cdot dA = EAcos \theta [/itex]; 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).
 

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