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
which is not the answer

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

How?Can you please elaborate.

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

Ok.Let me ask one thing
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?

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