The flux through the hemispherical surface

[gracy]

Homework Statement

The flux through the hemispherical surface in the figure shown below is

Homework Equations

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

 

[Hesch]

The flux = π*R² * E.

Think about what π*R² actually is.

 

[gracy]

Think about what π*R2 actually is.

area of a circle.

 

[haruspex]

the flux through the hemispherical surface should be =Q/2ε0
but this is wrong it should be πR^2E

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?

 

[gracy]

Do any flux lines pass through the hemispherical shell but not through the disk, or vice versa?

No.

 

[gracy]

What is the flux of E through the disk?

If we go for finding flux for disk
It will be E cos 180 degrees 4pir^2
=-4pir^2E
which is not the answer

 

[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. [edit] 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.

 

[Hesch]

What is the flux of E through the disk?

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: ψ_disk + ψ_hemis = 0 →
ψ_hemis = -ψ_disk

 

[gracy]

bowl

what's that?

 

[BvU]

The curved part of the hemisphere

 

[gracy]

The flux through the disk = πR2*E.

How?Can you please elaborate.

 

[BvU]

You said so yourself in #6 !

 

[gracy]

You said so yourself in #6 !

I wrote

-4pir^2E

 

[BvU]

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

Agrees nicely, I would say !

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

 

[Hesch]

How?Can you please elaborate.

Ψ = Area*Ψ_dens*sin φ = πR²*E*1.

How does 4πR² get into it ?

 

[haruspex]

If we go for finding flux for disk
It will be E cos 180 degrees 4pir^2
=-4pir^2E
which is not the answer

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

 

[gracy]

Where did you get that from

It was always there!

 

[gracy]

How do you get 4πR24\pi R^2? What's the area of a disk?

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

 

[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?

 

[haruspex]

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

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

 

[haruspex]

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?

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

 

[gracy]

what is difference between area and surface area?

 

[gracy]

difference between area and surface area?

Don't reply me.I got the answer.

 

[gracy]

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

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

 

[gracy]

What flux does that give?

-EπR2

 

[gracy]

When you close a volume by the disk+hemisphere, that does not include any charge

Then from where all these electric field lines are coming?Are these external electric fields?

 

[haruspex]

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

There is no dichotomy here. The two methods are completely consistent.
Say you have a cube with a charge somewhere inside it. You can use Gauss' law to find the total flux very easily; but if you were to figure out exactly what the field looks like, integrate the field over each of the six square surfaces to find the flux through them then add these up, you would get the same answer. Just use the method that suits the question.

In the present case, you are given a uniform field E passing through a hemispherical shell parallel to its axis. You do not know what is generating that field, or where it is - it doesn't matter as far as the flux through the hemispherical surface is concerned. By presuming that the field remains continuous over the whole interior of the hemisphere, and adding a disk surface to enclose the region, we can do two things:
- easily compute the flux entering the half sphere region through the disk;
- use Gauss' law to deduce that an equal flux passes out through the hemispherical surface of the region.

 

[gracy]

What is surface area of disk?

 

[haruspex]

What is surface area of disk?

You seem to be stuck on the concept of surface area as though it is something distinct from area of a plane figure.
Take any closed 3D region. It has a bounding surface. You can cut that surface up into a number of surfaces, none of which in itself bounds a volume, but each still has an area. You could call each a surface area since it is the area of a surface, just not a closed surface. E.g., the six faces of a cube. The surface area of the 3D region is simply the sum of the areas of the pieces.

 

[gracy]

The surface area of the 3D region is simply the sum of the areas of the pieces.

So,how surface area of sphere comes out to be 4 pi r^2?Does it include 4 circles?I don't think so