# Flux through surface of half a cone

1. Jan 22, 2007

### physgirl

1. The problem statement, all variables and given/known data
find electric flux that enters left hand side of a cone (with base radius R and height h). electric field penetrates the cone horizontally (cone is on a horizontal table).

2. Relevant equations
flux = surface integral of EdA
where E represents component of electric field normal to surface

3. The attempt at a solution
so that's the formula I got from my book... so I thought I had to find the integral of Ecos(theta)dA.... since Ecos(theta) would be the component of E that's normal to slanted surface of the cone... and if I integrate with respect to A I get Ecos(theta)A... cos(theta) is h/sqrt(h^2+R^2).... area of cone (minus the base) is pi*R*sqrt(h^2+R^2)...

so I did... E*(h/sqrt(h^2+R^2))*[pi*R*sqrt(h^2+R^2)]/2
once I cancel out stuff, I get: E*(h)(pi*R)/2... 2piERh
however, the back of my book says the answer is ERh... why??

2. Jan 22, 2007

### Gokul43201

Staff Emeritus
If theta is the semi-vertical angle (as it appears to be), then that is not the angle between the E field and the normal to each area element (it will only be true of those field lines that pass through the axis of the cone). Consider, for instance, a field line that just barely glances past the surface (i.e., is tangential to it) - clearly the angle between E and A is then 90*.

You need to first be clear about the angles involved before you can proceed. Does this make sense to you?

PS: Hint for proceeding - notice the relation between the dot product of two vectors and the projection of one vector along the other.

Last edited: Jan 22, 2007
3. Mar 8, 2007

### susuking

the reason why your ans is wrong is because the angle that you've calculated need some thinking, and think of the effective surface area facing the flux, knowing that, you should have your answer