Finding flux through a part of infinite disc

In summary: This question was from FIITJEE AITS part test II. Are you giving this exam ?No, I'm not familiar with that exam.In summary, two individuals discuss a problem involving finding the flux of a charge on an infinite disc using integration. One individual suggests using a small sphere to simplify the problem, while the other individual suggests using spherical caps. Eventually, both individuals reach the solution. One of the individuals mentions seeing similar examples before, and the other asks where they can find more examples.
  • #1
Saitama
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Homework Statement


A charge ##(5\sqrt{2}+2\sqrt{5})## coulomb is placed on the axis of an infinite disc at distance ##a## from the centre of disc. The flux of this charge on the part of the disc having inner and outer radius of ##a## and ##2a## will be

A)##\dfrac{3}{2\epsilon_0}##

B)##\dfrac{1}{2\epsilon_0}##

C)##\dfrac{2(\sqrt{5}+\sqrt{2})}{\epsilon_0}##

D)##\dfrac{2\sqrt{5}+5\sqrt{2}}{2\epsilon_0}##

Answer: A

Homework Equations


The Attempt at a Solution


I have solved the given problem through integration but since this is an exam problem, I am wondering if there is a smarter way to solve this problem. Since the disc is infinite, I think we can use this fact to find a shorter method. Or is integration the only way to solve the problem?

Any help is appreciated. Thanks!
 
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  • #2
Did you integrate by breaking the annular region of the disk into nested rings and integrating the normal component of E over the rings?

If so, there is a trick you can do to set up a much simpler integral that will yield the result. (You can avoid actually doing this integration if you happen to know the formula for the surface area of a spherical cap (hint). But that's something I would have to look up :blushing:)
 
  • #3
TSny said:
Did you integrate by breaking the annular region of the disk into nested rings and integrating the normal component of E over the rings?

Yes, that's exactly how I did it. :)

(You can avoid actually doing this integration if you happen to know the formula for the surface area of a spherical cap (hint). But that's something I would have to look up :blushing:)

I think I understand your trick but I don't seem to able to apply it. Don't we have two spherical caps here? One of radius ##\sqrt{2a^2}## and the other of ##\sqrt{5a^2}##.
 
  • #4
I was thinking of working with a single sphere which you could take to be a sphere of unit radius. But, yes, there will be two spherical caps on this sphere to consider, if you want to avoid integration. Or, you can set up a simple integral on this sphere and not worry about looking up a formula for the area of a spherical cap.
 
  • #5
TSny said:
I was thinking of working with a single sphere which you could take to be a sphere of unit radius. But, yes, there will be two spherical caps on this sphere to consider, if you want to avoid integration. Or, you can set up a simple integral on this sphere and not worry about looking up a formula for the area of a spherical cap.

Umm...I don't get it, I have attached a figure, do I have to find the flux through the red spherical cap? I am thinking that I will have to deal with solid angle.
 

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  • #6
Draw a small sphere around the charge with a radius considerably smaller than ##a##. You can think of it as having unit radius.

EDIT: Here's a pic
 

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  • Annular flux.png
    Annular flux.png
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  • #7
TSny said:
Draw a small sphere around the charge with a radius considerably smaller than ##a##. You can think of it as having unit radius.

EDIT: Here's a pic

Brilliant! That is such a nice method, thank you very much TSny! :)

How did you think of the sphere? And what's wrong with my spherical caps?
 
  • #8
Pranav-Arora said:
How did you think of the sphere? And what's wrong with my spherical caps?

I'm not sure how the thought came to me. Probably I've seen similar examples before. (I'm old)

Your spherical caps are just fine! It's a nice way and it will require about the same work as using a single sphere.
 
  • #9
Very sorry for the delay in reply, I did not have access to internet.

TSny said:
I'm not sure how the thought came to me. Probably I've seen similar examples before. (I'm old)

Can you please tell me where you saw these kind of examples? :)

Your spherical caps are just fine! It's a nice way and it will require about the same work as using a single sphere.
Yes, I reached the answer from my spherical caps too, thank you for the awesome solution TSny! :smile:
 
  • #10
Pranav-Arora said:
Very sorry for the delay in reply, I did not have access to internet.
Can you please tell me where you saw these kind of examples? :)Yes, I reached the answer from my spherical caps too, thank you for the awesome solution TSny! :smile:

This question was from FIITJEE AITS part test II. Are you giving this exam ? Same here. I was not able to solve this problem during exam. It was only at home that I solved it by manual integration.

But your method is awesome shortcut TSny ! Thanks ! :)
 
  • #11
Pranav-Arora said:
Can you please tell me where you saw these kind of examples? :)

I really can't say. You'll just pick up this sort of thing with practice.

Yes, I reached the answer from my spherical caps too

Good work!
 
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Related to Finding flux through a part of infinite disc

1. What is the formula for finding flux through a part of an infinite disc?

The formula for finding flux through a part of an infinite disc is Φ = 2πσR, where Φ is the flux, σ is the surface charge density, and R is the radius of the disc.

2. How is flux different from electric field?

Flux is a measure of the amount of electric field passing through a given area, while electric field is a measure of the force experienced by a charged particle in an electric field. Flux takes into account both the strength and direction of the electric field, while electric field only considers the strength.

3. Can flux through a part of an infinite disc be negative?

Yes, flux through a part of an infinite disc can be negative. This would occur if the electric field is pointing in the opposite direction of the normal vector of the surface, causing the flux to have a negative value.

4. How does the radius of the disc affect the flux?

The radius of the disc directly affects the flux through a part of the disc. As the radius increases, the surface area of the disc also increases, resulting in a larger area for the electric field to pass through. This leads to a higher flux value.

5. Can the flux through a part of an infinite disc be calculated using calculus?

Yes, the flux through a part of an infinite disc can be calculated using calculus. This involves integrating the electric field over the surface of the disc to determine the total flux passing through the area.

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