# Homework Help: The flux through the hemispherical surface

1. Sep 20, 2015

### gracy

1. The problem statement, all variables and given/known data
The flux through the hemispherical surface in the figure shown below is

2. Relevant equations

3. The attempt at a solution
But the above formula is for closed surface .But hemisphere is not a closed surface to make it closed surface we will need another hemisphere .
then the flux through the hemispherical surface should be =Q/2ε0
but this is wrong it should be πR^2E

2. Sep 20, 2015

### Hesch

The flux = π*R2 * E.

Think about what π*R2 actually is.

3. Sep 21, 2015

### gracy

area of a circle.

4. Sep 21, 2015

### haruspex

The answer must be expressed in terms of the facts given. Your diagram indicates a (uniform?) field and a radius. It does not show a charge.
To answer the question in terms of E, imagine closing the hemisphere with a flat disc radius R. What is the flux of E through the disk? Do any flux lines pass through the hemispherical shell but not through the disk, or vice versa?

5. Sep 21, 2015

### gracy

No.

6. Sep 21, 2015

### gracy

If we go for finding flux for disk
It will be E cos 180 degrees 4pir^2
=-4pir^2E

7. Sep 21, 2015

### BvU

"which is not the answer" is not correct: Haru asks for the flux through the disk and that is $-4\pi r^2\,E\$.

1.  sorry, overlooked the 4. flux through the disk and that is $\ -\pi r^2\,E\$.

You did very well letting the area vector for the disk point to the left (the outside of the hemisphere).

Now your relevant equation comes in very handy: Let S be the hemisphere. There is no charge inside, so $\oint \vec E \cdot d\vec S = 0$. For the disk part you have a value and the bowl has to have the equal and opposite value to get that zero.

Last edited: Sep 21, 2015
8. Sep 21, 2015

### Hesch

The flux through the disk = πR2*E.

The E-field is perpendicular to the disk ( not to the all over hemisphere ).

When you close a volume by the disk+hemisphere, that does not include any charge, the sum of the fluxes: ψdisk + ψhemis = 0 →
ψhemis = -ψdisk

9. Sep 21, 2015

### gracy

what's that?

10. Sep 21, 2015

### BvU

The curved part of the hemisphere

11. Sep 21, 2015

### gracy

12. Sep 21, 2015

### BvU

You said so yourself in #6 !

13. Sep 21, 2015

### gracy

I wrote

14. Sep 21, 2015

### BvU

Agrees nicely, I would say !

 Wow, now I see a 4. Where did you get that from

15. Sep 21, 2015

### Hesch

Ψ = Area*Ψdens*sin φ = πR2*E*1.

How does 4πR2 get into it ?

16. Sep 21, 2015

### haruspex

How do you get $4\pi R^2$? What's the area of a disk?

17. Sep 21, 2015

### gracy

It was always there!

18. Sep 21, 2015

### gracy

I thought we use surface area in gauss's formula,not area.

19. Sep 21, 2015

### gracy

surface area is always of three dimensional figures?it can not be there in two dimensional figures such as circle and square?

20. Sep 21, 2015

### haruspex

You have a uniform field E at right angles to a plane surface (disk) of area $\pi R^2$. What flux does that give?

21. Sep 21, 2015

### haruspex

Disks and squares have areas. They might or might not happen to be part of the surface of a three dimensional shape.

22. Sep 21, 2015

### gracy

what is difference between area and surface area?

23. Sep 21, 2015

### gracy

24. Sep 21, 2015

### gracy

So we will take surface area in gauss's law if there is any 3d figure otherwise we will use formula of area.

Last edited: Sep 21, 2015
25. Sep 21, 2015

-EπR2