# Flux in a cone

1. May 24, 2010

### jyothsna pb

1. The problem statement, all variables and given/known data

what is the flux entering a cone of radius r and height h located in a uniform electric field E parallel to its base?

2. Relevant equations

flux = E.ds

3. The attempt at a solution

2. May 24, 2010

### ehild

What is ds?

ehild

3. May 24, 2010

### gabbagabbahey

Better yet, what does the divergence theorem tell you?

4. May 24, 2010

### jyothsna pb

it is the small element of the effective area through which flux passes

5. May 24, 2010

### gabbagabbahey

He was hinting that your first step should be to find an expression for ds (if you evaluate the flux integral directly).

6. May 24, 2010

### ehild

And what is the effective area of a curved surface like that of a cone?

ehild

7. May 24, 2010

### jyothsna pb

dat is what i am confused of

8. May 24, 2010

### jyothsna pb

i see 2 possibilities one the curved surface area of the cone and the other the triangular plane perpendicular to base but which one?

9. May 24, 2010

### ehild

In your first formula dΦ= E˙ds is the product of E with the area normal to it. As the field lines are parallel to the base, these surface elements are parts of that triangular cross section of the cone which is perpendicular to the electric field. As E is constant simply multiply E with the area of the triangle.

10. May 24, 2010

### jyothsna pb

thanx but one more doubt cant we take curved surface area resolved into the field direction in this case?

11. May 24, 2010

### ehild

I do not quite follow you, but this is what we do. By definition, the flux is E ds cos (alpha) where alpha is the angle between E and the normal of the surface element. ds cos(alpha) is the projection of the surface area onto a plane normal to E.

ehild

Last edited: Jun 29, 2010
12. May 24, 2010

### jyothsna pb

thanku so much got the point now