Calculate: the net electric flux = flux through rounded portion + flux

In summary, Gauss's law is used to calculate the net electric flux, which is equal to the flux through the rounded portion plus the flux through the flat base. The flux through the flat base is negative, meaning it is directed into the object. This is because the total flux through the whole hemisphere is a positive value, so if the flux through the rounded part is larger than that value, the flux through the flat part must be negative. This is all based on the equation: \Phi_{E}=\oint_{S} \vec{E} \bullet d\vec{S}, where S is any closed surface. The second question is a bit tricky, but it can be explained by considering the distribution of charges and the
  • #1
cuongbui1702
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testbank-27-chap23_zpsc5494d24.png

I use Gauss's law to calculate: the net electric flux = flux through rounded portion + flux through the flat base = Q/ε. And the flux through the flat base is negative, the negative means?? Help me please, it is very confusing

2)
testbank-27-chap23_zps908a0331.png

Why they choose C, i think E is correct because Q doesnot identify.
 
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  • #2
Who help mẹ please :(
 
  • #3
Well I guess for the second question it can be because if the charge Q is strong enough it can re-arrange electrons inside the material of the neutral to some extent.
 
  • #4
And for the first part, it's because your total flux through the whole hemisphere is roughly 7.45[itex]\times10^{4}[/itex] ,so if the flux through the rounded part is more than that value, the flux through the flat part has to be a negative amount.
 
  • #5
Remember this equation, it's quite helpful:

[itex]\Phi_{E}=\oint_{S} \vec{E} \bullet d\vec{S}[/itex]

where S is any closed surface
 
  • #6
TheAustrian said:
And for the first part, it's because your total flux through the whole hemisphere is roughly 7.45[itex]\times10^{4}[/itex] ,so if the flux through the rounded part is more than that value, the flux through the flat part has to be a negative amount.
I think it is all positive charge so that E is out of it=> no flux through in=> don't exist? why I'm wrong
 
  • #7
cuongbui1702 said:
I think it is all positive charge so that E is out of it=> no flux through in=> don't exist? why I'm wrong

because you are told in the question how the flux is directed. I am not sure how else to explain it, but with the equation stated above your last post, you should be able to calculate your total flux out of the whole geometrical object, which is 7.45×104. So if one portion of the geometrical object has more flux than that value, then the other portion must have a negative amount of flux. (Negative flux means flux into the object, positive flux means directed out from the object)
 
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  • #8
A) The total charge is given, not the distribution. Imagine a dipole +1 microcoulomb close to the top of the hemishpere and -1/3 close to the bottom plate. Fieldlines go out at the top, so positive flux, in at the bottom, negative flux.

B) A bit of a trick question. The field inside the conductor is zero. There are two contribuants: Q and the induced surface charge. First one points away from Q, so the other has to point towards Q.
But I have great difficulty with this answer, since nor Q nor the charges on the surface really produce a field inside the conductor.
 
  • #9
BvU said:
But I have great difficulty with this answer, since nor Q nor the charges on the surface really produce a field inside the conductor.

Why that?
 
  • #10
One can look at the field inside the conductor (Yes. It's zero. ) as the superposition of two fields. One field is due to charge Q. The other is due to surface charge on the conducting sheet. The field due to the surface is directed opposite the field due to the Charge Q. However, the problem does not state the sign of charge Q, so we really can't say whether the field due to the surface is directed towards Charge Q or directed away from charge Q.
 

1. What is electric flux?

Electric flux is a measure of the amount of electric field passing through a given surface. It is represented by the symbol Φ and is measured in units of volts per meter (V/m).

2. How is electric flux calculated?

To calculate electric flux, you need to first determine the electric field strength at each point on the surface. Then, you multiply the electric field strength by the area of each small section of the surface and add them all together. This will give you the total electric flux through the surface.

3. What is the significance of the rounded portion in the formula for net electric flux?

The rounded portion in the formula for net electric flux represents the contribution of any curved or rounded surfaces to the total flux. This is important because the electric field lines passing through a curved surface may not be perpendicular to that surface, so the calculation of flux must take this into account.

4. Can the net electric flux ever be negative?

Yes, the net electric flux can be negative. This occurs when the electric field lines are leaving the surface, indicating that the net flow of electric field is outward from the surface. In this case, the electric flux is represented by a negative value.

5. What factors can affect the net electric flux through a surface?

The net electric flux through a surface can be affected by the strength and direction of the electric field, the size and shape of the surface, and the angle between the electric field lines and the surface. Additionally, the presence of other charges or conducting materials nearby can also influence the net electric flux.

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