# Calculate the flux through the surface?

1. Feb 26, 2013

### ehabmozart

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

A hemispherical surface with radius e in a region of uniform electric field E has its axis aligned parallel to the direction of the field. Calculate the flux through the surface?

2. Relevant equations

Flux = E A cos β

3. The attempt at a solution

Why can't we say that the flux is E * Area of hemisphere which is E (2∏r^2)+(∏r^2) ???

Thanks for clarifying!

2. Feb 26, 2013

### Staff: Mentor

Do the quantities either side of your plus sign have identical units??

That aside, if you want to involve a constant flux and the area of a hemisphere then you'd need that the flux be uniform over that area. But is isn't. Some of the hemisphere's surface is almost parallel to the E field, while some is almost perpendicular to E.

BTW, I think the hemisphere in question is 'open to the atmosphere', a surface comprising half a spherical shell.

3. Feb 26, 2013

### ehabmozart

Allright, I knew that the answer should be E (pi r^2) only... They said it should involve the down circle only... But my question is why? ... And a reply to your is that yes, the electric field should be constant over a hemisphere since the radius is the same everywhere... Hence, the flux should so be the same everywhere!!

4. Feb 26, 2013

### tiny-tim

hi ehabmozart!

hint: suppose it was a closed surface (the hemisphere plus the base) …

what would the flux be then?

5. Feb 26, 2013

### ehabmozart

Hi tinytim! :)

I guess your hint was the question itself... I consider the answer to be E * (2 pi r^2 + pi r^2) and not as the book suggests E pi r^2 :) Kindly help!

6. Feb 26, 2013

### tiny-tim

hmm …

how generally would you calculate the flux through a closed surface?

any theorems?

7. Feb 26, 2013

### ehabmozart

All what i've take is that Flux = E A ... We didn't even reach gauss's law so i want an easy approach to this!

8. Feb 26, 2013

### tiny-tim

9. Feb 26, 2013

### ehabmozart

can i get more clarification on why the electric flux through the circle will be the same which passes through the top part of the hemisphere?

10. Feb 26, 2013

### haruspex

Since the field is uniform and normal to the circle, every bundle of flux that passes through the hemispherical shell will pass through the circle, and vice versa.

11. Feb 26, 2013

### ehabmozart

Ok... that's fine.. But y?? I mean the circle is not having the same surface area as the hemisphere does?

12. Feb 26, 2013

### Staff: Mentor

Sketch a field of parallel flux lines, and draw in a hemisphere. (No need to post it here, so don't worry about not having good sketching ability.)

Count the number of flux lines that pass through the hemispheric surface. Compare this count with the number of flux line that pass through the flat circular surface. Can you see any relationship?

13. Feb 27, 2013

### tiny-tim

(just got up :zzz:)
hint: what is the divergence (at a general point (x,y,z)) of an electric field of constant magnitude E in the z direction?

14. Jan 31, 2014

### BwriiBwear

The reason that it is E*(pi*r^2) isn't because only the bottom part matters. It's because, in the top part, the x-and y-components of flux cancel out, due to the n^ vectors coming out of the surface at varying directions, leaving only the contributions of the positive z-direction n^ vectors, and in the bottom, all n^ vectors point in the -z-direction. Thus, the net of the flux would be E(surface area of top part)-E(surface area of base)=E(2*pi*r^2)-(pi*r^2), leaving just E(pi*r^2).

15. Sep 15, 2017

### Fab

Since the E field goes inward on the circular base, the flux is -E(pi*r^2). Since it is pointing outward from the concaved part, the flux is E(2pi*r^2) (since it is half a sphere the area is halfed). Then just compine the two