Electric Flux through a semi-spherical bowl from a charged particle

AI Thread Summary
The discussion centers on calculating the electric flux through a semi-spherical bowl due to a charged particle, with a focus on using Gauss's Law. Participants express uncertainty about finding the surface area of the hemisphere and the implications of the bowl's material properties on the calculations. There is a debate over whether the dielectric constant affects the symmetry needed for applying Gauss's Law effectively. One participant calculates a flux value of 504, which is noted to be double the expected answer, leading to further clarification on accounting for the hemisphere's surface area. The conversation highlights the importance of precise problem statements and the relevance of dielectric properties in electric field calculations.
flamebane
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Homework Statement
A particle with a net charge of 4.46nC is suspended over a semi-spherical bowl having an inner radius of 10.6cm , exactly at the level of the bowl's rim. What is the electric flux across the bowl's inside surface? Assume e = 8.85*10^-12. Answer in (Nm^2)/C.
Relevant Equations
EACostheta
I believe this does has a couple of Calculus aspects to it but I don't really know how I'd find the surface area of inside the bowl.
The answer sheet says the answer is 252 with a margin of error of +/- 1
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Welcome to PF.

Is the particle at the center of the bowl's opening (to place it at the center of curvature of the bowl)? If so, there is a symmetry that helps to make the problem easier.

What equations do you know to calculate the electric flux from a charged particle?
 
berkeman said:
Welcome to PF.

Is the particle at the center of the bowl's opening (to place it at the center of curvature of the bowl)? If so, there is a symmetry that helps to make the problem easier.

What equations do you know to calculate the electric flux from a charged particle?
I would assume that the particle would be at the center suspended above, and as for equations theres: Φ=EA, Φ=EAcosΘ, for flat surfaces theres σ/2e0, E=λ/2πe0r, and I forgot the electric flux for a closed.
 
flamebane said:
I don't really know how I'd find the surface area of inside the bowl.
You don't know how to find the surface area of a hemisphere?
 
flamebane said:
I forgot the electric flux for a closed.
Do a quick Google search on Gauss' Law and click on the Wikipedia link (it's the first hit on the search list for me). Do you see anything in that article that would help you to answer this question? :smile:
 
berkeman said:
Do a quick Google search on Gauss' Law and click on the Wikipedia link (it's the first hit on the search list for me). Do you see anything in that article that would help you to answer this question? :smile:
Q/e0 right? Q would be net charge which would be 4.46*10^-9 and e0 is stated in the question?
 
flamebane said:
Q/e0 right? Q would be net charge which would be 4.46*10^-9 and e0 is stated in the question?
Don't forget that they are only asking for the flux through the hemisphere... :wink:

And I'm bothered that they did not specify the value of ##\epsilon_r## for the hemisphere/bowl material. Can you say why that would bother me?

flamebane said:
Assume e = 8.85*10^-12
And what does that mean? I'm pretty sure they did not mean to redefine the charge on an electron. Are they saying to assume that ##\epsilon_r = \epsilon_0## for the bowl, or just the space around the bowl?
 
berkeman said:
Don't forget that they are only asking for the flux through the hemisphere... :wink:

And I'm bothered that they did not specify the value of ##\epsilon_r## for the hemisphere/bowl material. Can you say why that would bother me? And what does that mean? I'm pretty sure they did not mean to redefine the charge on an electron. Are they saying to assume that ##\epsilon_r = \epsilon_0## for the bowl, or just the space around the bowl?
So does that mean I cant use Gauss's Law with Q/e0 ?
 
flamebane said:
So does that mean I cant use Gauss's Law with Q/e0 ?
No, you certainly can use Gauss' Law. And if they are specifying in the problem that the material of the bowl has ##\epsilon_r = \epsilon_0## then you only have to take into account that the surface area of the hemisphere is a fraction of the full sphere enclosing that charge. Can you post the full problem statement? You can upload an image of the problem statement by using the "Attach files" link below the Edit window.
 
  • #10
berkeman said:
No, you certainly can use Gauss' Law. And if they are specifying in the problem that the material of the bowl has ##\epsilon_r = \epsilon_0## then you only have to take into account that the surface area of the hemisphere is a fraction of the full sphere enclosing that charge. Can you post the full problem statement? You can upload an image of the problem statement by using the "Attach files" link below the Edit window.
The File that I uploaded is the full problem statement
 
  • #11
flamebane said:
The File that I uploaded is the full problem statement
Okay. Not your fault, but the problem statement is not very good, since any bowl material will interact with the fields, but whatever.

So if you assume that ##\epsilon_r = \epsilon_0## that they give, and factor in the hemisphere versus full sphere/surface, what do you get for an answer?
 
  • #12
berkeman said:
Okay. Not your fault, but the problem statement is not very good, since any bowl material will interact with the fields, but whatever.

So if you assume that ##\epsilon_r = \epsilon_0## that they give, and factor in the hemisphere versus full sphere/surface, what do you get for an answer?
Using Gauss's Law I got 504, which is double what it should be
 
  • #13
flamebane said:
Using Gauss's Law I got 504, which is double what it should be
Did you divide by two to account for the hemisphere? Can you please show your calculation? Thanks.
 
  • #14
I converted the net charge from nC to C and then divided it by ε
berkeman said:
Did you divide by two to account for the hemisphere? Can you please show your calculation? Thanks.
 
  • #15
That's no help. It still looks like you just used the flux equation for a full sphere enclosing the charge. They are asking about the flux through a hemisphere.
 
  • #16
For whatever it's worth, the normal component of the electric field vector ##\mathbf{E}## is discontinuous across the surface of the bowl whilst the normal component of the field displacement vector ##\mathbf{D}## is continuous. However, the problem is clearly asking to find the flux "across the bowl's inside surface" which I interpret to mean that one has to draw a Gaussian surface just inside the skin of the bowl where there is no electric filed discontinuity to worry about. The dielectric constant of the bowl is not relevant.
 
  • #17
kuruman said:
The dielectric constant of the bowl is not relevant.
I believe that is not correct. If the hemisphere permittivity differs from vacuum , then one can no longer use spherical symmetry arguments to adduce the electric field. Gauss still holds but the symmetry does not.
EDIT: I no longer believe this is correct. The fields do remain radial as described by @kuruman. Apologies it is late
 
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