Is Electric Flux Through a Half Sphere Zero According to Gauss' Law?

[MatinSAR]

Homework Statement

Find the electric flux passing through surface 1 and surface 2.

Relevant Equations

Gauss's Law

Picture :

My answer :
I guess net electric flux is 0.
so electric flux passing through surface 1 = -(electric flux passing through surface 2)
and electric flux passing through surface 1 is EA = E(pi)(r^2)

Is it correct? Thank you ...

 

[vanhees71]

Is the electric field explicitly given to be homogeneous? Then you are right.

 

[MatinSAR]

Is the electric field explicitly given to be homogeneous? Then you are right.

Do you mean a uniform field ?
According to the picture it's uniform.
Thank you.

 

[vanhees71]

Yes, $\vec{E}=\text{const}$. Then it's correct, but I guess you should explicitly prove it. There are in principle two ways: The simple one is using Gauss's integral law,
$$\int_V \mathrm{d}^3 x \vec{\nabla} \cdot \vec{E}=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E},$$
with which you are immediately done, or you explicitly calculate the fluxes through the two surfaces.

 

[Orodruin]

Unless the question is to find the flux passing the surfaces 1 and 2 individually. The wording is a bit ambiguous as many times will be the case in ordinary language.

 

[MatinSAR]

Unless the question is to find the flux passing the surfaces 1 and 2 individually.

Yes this was what I meant.(find the flux passing the surfaces 1 and 2 individually)
So my answer is true, right?

Yes,

Thank you but I have not read those equations yet ...
This is my book : Fundamentals of Physics 10th Edition

 

[PhDeezNutz]

@vanhees71 How does the Divergence Theorem/Gauss’ Law work for (two) non-closed surfaces?

Am I missing something here? Unless I’m having a complete brain fart I think we have to calculate the flux explicitly.

Then again, OP says he’s not familiar with the Divergence Theorem so maybe he’s not familiar with explicitly calculating surface integrals. So maybe the expectation is to heuristically conclude that “every field line that enters the circle also leaves the hemisphere so therefore the flux through each surface is the same”

It’s always tricky when dealing with introductory courses.

 

[hutchphd]

Am I missing something here? Unless I’m having a complete brain fart I think we have to calculate the flux explicitly.

I hate to tell you but (flat circle) + (half spherical shell) is a closed surface. (You are excused.)

 

[PhDeezNutz]

I hate to tell you but (flat circle) + (half spherical shell) is a closed surface. (You are excused.)

I interpreted the question as find the flux through each of the surfaces independently. Not the combined surface.

But like @Orodruin said the wording is ambiguous.

 

[nasu]

This is what the OP did. He calculated the flux through the flat surface explicitly and only after this he used Gauss' law to say that the flux through the half sphere is the same in magnitude.

 

[hutchphd]

This is a bad bit of mental flatulence!
A very useful technique so be sure you see it.

 

[PhDeezNutz]

This is what the OP did. He calculated the flux through the flat surface explicitly and only after this he used Gauss' law to say that the flux through the half sphere is the same in magnitude.

In that case @vanhees71 ‘s post makes a lot more sense and I stand corrected.

1) Conclude that the flux through the circle is $E \pi r^2$

2) From Gauss’ Law conclude that the net flux through the the combined surface is zero since there is no net charge enclosed. This is where the suggestion to use the Divergence Theorem for Explicit calculation came in because the integrand is zero and so then the integral must be zero.

3) From there conclude that the flux through the hemisphere must also be $E \pi r^2$

I’m losing it haha.

Christ I am so embarrassed.

 

[Orodruin]

Funny note: You can also apply Stokes’ theorem on this question. You just have to find a field whose curl is constant, which is not very difficult, and then you can conclude that the integrals must be the same because the boundary curve is the same.

Edit: Update, my brain is a bit slow after half a year of paternity leave ... you do not even need to find the field whose curl is constant, just know that it exists - which is clear from the fact that the constant field is divergence free.

 

[MatinSAR]

Homework Statement:: Find the electric flux passing through surface 1 and surface 2.
Relevant Equations:: Gauss's Law

Picture :
View attachment 298393
My answer :
I guess net electric flux is 0.
so electric flux passing through surface 1 = -(electric flux passing through surface 2)
and electric flux passing through surface 1 is EA = E(pi)(r^2)

Is it correct? Thank you ...

Hi ... I guess there is a mistake ...
I think electric flux passing through surface 1 is not EA=E(pi)(r^2) and it's negative ...

Thanks to everyone for the help.

 

[hutchphd]

Tradtionally from a closed surface the positive flux direction is outward so flux 2 would usually be called the positive one. The value seems correct to me. Why do you think it incorrect?

 

[Orodruin]

Hi ... I guess there is a mistake ...
I think electric flux passing through surface 1 is not EA=E(pi)(r^2) and it's negative ...

Thanks to everyone for the help.

Whether it is positive or negative depends on the normal direction.

Tradtionally from a closed surface the positive flux direction is outward so flux 2 would usually be called the positive one. The value seems correct to me. Why do you think it incorrect?

However, the OP’s interpretation is to find the flux through each surface, not the closed surface. With the normal of surface 1 upward, the closed surface integral with outward normal is (2)-(1) (with (k) representing the integral over surface k) and so, since divergence is zero, (2) = (1).

 

[hutchphd]

However, the OP’s interpretation is to find the flux through each surface, not the closed surface. With the normal of surface 1 upward, the closed surface integral with outward normal is (2)-(1) (with (k) representing the integral over surface k) and so, since divergence is zero, (2) = (1).

Why would that be chosen as positive? Without a definition the sign of the flux is arbitrary. No coordinates are defined.
So I was suggesting the closed surface definition (this is, after all, a Gauss Law question at heart)

 

[MatinSAR]

Tradtionally from a closed surface the positive flux direction is outward so flux 2 would usually be called the positive one. The value seems correct to me. Why do you think it incorrect?

Yes flux 2 would be called the positive one. At first I said the positive one is flux 1 and it's false.

Whether it is positive or negative depends on the normal direction.

Yes it depends on normal direction and here angle between normal and E is 180 so it's negative ...

 

[Orodruin]

Yes it depends on normal direction and here angle between normal and E is 180 so it's negative ...

I disagree. You have not been given a normal direction.

 

[kuruman]

I agree that the task here is to calculate the flux through two separate surfaces:
1. One surface is hemispherical and is bound by a closed loop of radius $r$.
2. The second surface is planar and is also bound by a closed loop of radius $r$.
To find the flux through each surface one has to
(a) orient the loop which in turn defines the normal to the surface or define the normal and then orient the loop. The right hand rule is to be used.
(b) Do the integral $\Phi_i=\int_{s_i}\vec E\cdot \hat n_i~dA~~(i=1,2)~$ for each surface. Depending on how the normals are defined, the fluxes may be equal or they may have opposite signs but their magnitudes will be equal.

Personally, I think the point of this exercise is to do the integral over the hemisphere explicitly without using the Gauss's law shortcut. A further generalization would be that if the bounding loop were kept fixed but the surface were distorted beyond a hemisphere and made to look like a chef's hat for example, the flux would still be the same.

 

[hutchphd]

Personally, I think the point of this exercise is to do the integral over the hemisphere explicitly without using the Gauss's law shortcut.

Perhaps, but it is introductory physics and the title of the thread is

"Gauss' Law in half sphere"