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

In summary, the problem asks for the electric flux through a hemisphere filled with a charge, but does not specify the value of ##\epsilon_r##. Using Gauss's Law, I got the flux to be double what it should be.
  • #1
flamebane
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1
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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  • #2
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?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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:
 
  • #6
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?
 
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  • #7
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?
 
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  • #8
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 ?
 
  • #9
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?
 
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  • #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.
 
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  • #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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FAQ: Electric Flux through a semi-spherical bowl from a charged particle

What is electric flux?

Electric flux is a measure of the electric field passing through a given surface. It is mathematically defined as the dot product of the electric field and the area vector of the surface. The SI unit of electric flux is the volt-meter (V·m).

How do you calculate the electric flux through a semi-spherical bowl from a charged particle?

To calculate the electric flux through a semi-spherical bowl from a charged particle, you can use Gauss's law. For a point charge \( q \) located at the center of an imaginary sphere, the total electric flux through the sphere is \( \frac{q}{\epsilon_0} \). Since the semi-spherical bowl is half of this sphere, the electric flux through the bowl is \( \frac{q}{2\epsilon_0} \).

Does the position of the charged particle affect the electric flux through the semi-spherical bowl?

Yes, the position of the charged particle significantly affects the electric flux through the semi-spherical bowl. If the charged particle is not at the center of the semi-spherical bowl, you would need to integrate the electric field over the surface of the bowl to find the total flux, which can be more complex.

What role does the permittivity of free space (\( \epsilon_0 \)) play in calculating electric flux?

The permittivity of free space (\( \epsilon_0 \)) is a constant that appears in Gauss's law and relates the electric flux through a closed surface to the charge enclosed by that surface. It essentially scales the amount of flux for a given charge. In SI units, \( \epsilon_0 \) is approximately \( 8.854 \times 10^{-12} \, \text{F/m} \) (farads per meter).

Can electric flux be negative?

Yes, electric flux can be negative. The sign of the electric flux depends on the direction of the electric field relative to the orientation of the surface area vector. If the electric field lines are entering the surface, the flux is considered negative; if they are exiting the surface, the flux is considered positive.

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