Gauss' Law in half sphere

In summary, the electric flux passing through surface 1 is not the same as the electric flux passing through surface 2.
  • #1
MatinSAR
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Homework Statement
Find the electric flux passing through surface 1 and surface 2.
Relevant Equations
Gauss's Law
Picture :
1647349429249.png

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 ...
 
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  • #2
Is the electric field explicitly given to be homogeneous? Then you are right.
 
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  • #3
vanhees71 said:
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.
 
  • #4
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.
 
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  • #5
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.
 
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  • #6
Orodruin said:
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?
vanhees71 said:
Yes,
Thank you but I have not read those equations yet ...
This is my book : Fundamentals of Physics 10th Edition
 
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  • #7
@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.
 
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  • #8
PhDeezNutz said:
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.:smile:)
 
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  • #9
hutchphd said:
I hate to tell you but (flat circle) + (half spherical shell) is a closed surface. (You are excused.:smile:)
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.
 
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  • #10
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 throught the half sphere is the same in magnitude.
 
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  • #11
This is a bad bit of mental flatulence!
A very useful technique so be sure you see it.
 
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  • #12
nasu said:
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 throught 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.
 
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  • #13
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.
 
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  • #14
MatinSAR said:
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. 🙏🙏🙏
 
  • #15
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?
 
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  • #16
MatinSAR said:
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.

hutchphd said:
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).
 
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  • #17
Orodruin said:
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)
 
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  • #18
hutchphd said:
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.
Orodruin said:
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 ...
 
  • #19
MatinSAR said:
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.
 
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  • #20
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.
 
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  • #21
kuruman said:
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"
 
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1. What is Gauss' Law in a half sphere?

Gauss' Law in a half sphere is a mathematical equation that relates the electric flux through a closed surface to the charge enclosed by that surface. It states that the electric flux through a half sphere is proportional to the charge enclosed by the half sphere.

2. How is Gauss' Law used in a half sphere?

Gauss' Law in a half sphere is used to determine the electric field at any point on the surface of a half sphere, given the charge enclosed by the half sphere. It is also used to calculate the total electric flux through the half sphere.

3. What is the formula for Gauss' Law in a half sphere?

The formula for Gauss' Law in a half sphere is Φ = Q/ε0, where Φ is the electric flux, Q is the charge enclosed by the half sphere, and ε0 is the permittivity of free space.

4. What are the assumptions made in Gauss' Law for a half sphere?

The main assumptions made in Gauss' Law for a half sphere are that the electric field is constant on the surface of the half sphere and that the half sphere is a closed surface. Additionally, the charge enclosed by the half sphere must be point-like or distributed evenly throughout the enclosed volume.

5. How is Gauss' Law in a half sphere different from Gauss' Law in a full sphere?

Gauss' Law in a half sphere only considers the electric flux through half of a sphere, while Gauss' Law in a full sphere considers the electric flux through the entire surface of a sphere. Additionally, the charge enclosed in a half sphere is only half of the charge enclosed in a full sphere. However, the formula and assumptions for both laws are the same.

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