Potential due to induced charges

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The discussion revolves around calculating the electric potential at the center of a spherical shell when a point charge is placed inside it. Participants clarify that the shell is assumed to be conducting, leading to induced charges on both the inner and outer surfaces. The inner surface acquires a charge of -q, while the outer surface gains a charge of +q, affecting the potential calculation. It is emphasized that the potential at the center remains the same regardless of whether the charge distribution on the inner surface is uniform or non-uniform. Ultimately, the potential is derived from the contributions of the induced charges and the point charge, confirming that the potential at the center can be calculated effectively.
Saitama
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Homework Statement


A point charge q is placed at a distance ##r## from the centre O of a uncharged spherical shell of inner radius ##R## and outer radius ##2R##. The distance ##r<R##. Find the electric potential at the centre of shell.

Homework Equations


The Attempt at a Solution


The charges will be induced as shown in attachment. The problem is calculating the potential. Do I have to consider each surface of both the spheres and calculate the potential? But that would give ##kq/r## which is wrong. :confused:
 

Attachments

  • spheres with charges.png
    spheres with charges.png
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I don't quite understand your diagram. Looks to me like you have too many q's. There will be some charge induced on the inner surface of the shell and charge induced on the outer surface of the shell. Can you state the total amount of charge induced on each surface? Can you describe qualitatively how the charge will be distributed on each surface?
 
TSny said:
I don't quite understand your diagram. Looks to me like you have too many q's. There will be some charge induced on the inner surface of the shell and charge induced on the outer surface of the shell. Can you state the total amount of charge induced on each surface? Can you describe qualitatively how the charge will be distributed on each surface?

[strike]What's wrong with the diagram? Won't the induced charges be equal to q in magnitude? [/strike] :confused:

The q's near the surface are the induced ones. The one near the centre O is stated in the problem.

Ah yes, the shells are not given to be conducting. Is my diagram correct for the case when the shells are connecting?

But since the shells are not conducting, I don't know how the charge will be distributed. :(
 
Hello Pranav

Can you provide the answer to the problem ?
 
Tanya Sharma said:
Hello Pranav

Can you provide the answer to the problem ?

As per the answer key, it is ##kQ(1/r-1/(2R))##.
 
The answer corresponds to a conducting shell. Note that there is only one shell. It has an inner radius R and an outer radius 2R.
 
TSny said:
The answer corresponds to a conducting shell. Note that there is only one shell. It has an inner radius R and an outer radius 2R.

Sorry, I should have read the question properly. :redface:

Now it is easy to answer if it is given that the shells are conducting, thank you! :smile:

So the question is wrong as it isn't mentioned that the shells are conducting?

And what if there were two separate shells?
 
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Okay...

There are only two spherical surfaces not four as indicated by your diagram .

The shell is assumed to be conducting in this problem.

The charge induced on the inner surface(R) will be -q and that induced on the outer surface surface(2R) will be +q .

Now replace the metal with two spherical charge distribution .One with radius R and charge -q and other with radius 2R and charge +q .

Now in order to calculate potential at the center ,we have to find sum of potentials due to three charge distributions . +q(at distance r) , -q (sphere of charge of R radius ) ,+q(sphere of charge of 2R radius) .

You very well know potential due to a point charge.

What is the potential inside a thin spherical shell ?

Plug in the values and you get the answer .

Hope that helps .
 
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Tanya Sharma said:
Okay...

There are only two spherical surfaces not four as indicated by your diagram .

The shell is assumed to be conducting in this problem.

The charge induced on the inner surface(R) will be -q and that induced on the outer surface surface(2R) will be +q .

Now replace the metal with two spherical charge distribution .One with radius R and charge -q and other with radius 2R and charge +q .

Now in order to calculate potential at the center ,we have to find sum of potentials due to three charge distributions . +q(at distance r) , -q (sphere of charge of R radius ) ,+q(sphere of charge of 2R radius) .

You very well know potential due to a point charge.

What is the potential inside a thin spherical shell ?

Plug in the values and you get the answer .

Hope that helps .

Thanks Tanya but you were a bit late, TSny cleared it up. :)
 
  • #10
You have to be careful. Is the negative charge induced on the inner surface of the shell uniformly distributed? If not, how do you find the potential at the center from this non-uniformly distributed charge?
 
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  • #11
TSny said:
You have to be careful. Is the negative charge induced on the inner surface of the shell uniformly distributed? If not, how do you find the potential at the center from this non-uniformly distributed charge?

If it is non-conducting, no, they are not uniformly distributed. The negative charge induced is equal in magnitude to q. The distance of the induced charges is at equal distance from O so is the potential again##-kq/R## due to the inner shell?
 
  • #12
The induced charge will be uniform on the outer surface, but non-uniform on the inner surface.
 

Attachments

  • pt charge in shell.png
    pt charge in shell.png
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  • pt charge in shell 2.png
    pt charge in shell 2.png
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  • #13
TSny said:
The induced charge will be uniform on the outer surface, but non-uniform on the inner surface.

We are talking about the non-conducting case, right?
 
  • #14
I was assuming a conducting shell. I think that's what leads to the answer you gave. If it's non-conducting, then I believe it would be fairly complicated. The induced charge would depend on the dielectric constant of the material and the surfaces of the shell would not be equipotential surfaces.
 
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  • #15
TSny said:
I was assuming a conducting shell. I think that's what leads to the answer you gave. If it's non-conducting, then I believe it would be fairly complicated. The induced charge would depend on the dielectric constant of the material and you would the surfaces of the shell would not be equipotential surfaces.

Thanks TSny for the explanation. :smile:

Can you give me a link which proves why the charges arrange like you have shown in your attachment? I did learn this a year ago but I seem to have forgotten now. :)
 
  • #16
Pranav-Arora said:
Can you give me a link which proves why the charges arrange like you have shown in your attachment? I did learn this a year ago but I seem to have forgotten now. :)

I don't know of a link offhand. One way to look at is as follows. There's a "uniqueness theorem" of electrostatics that states that if you can find any solution of your problem that satisfies all the boundary conditions, then it will be the only possible solution. Our boundary conditions here are that all points of the conductor must be at the same potential and the net charge of the conductor is zero. Also, the potential must go to zero at infinity.

Consider the region of space outside the shell. This is a region of space bounded by the outer surface of the conductor and infinity. The outer surface of the conductor is an equipotential spherical surface and also the potential goes to zero at infinity. It can be shown that any solution to Laplace's equation for the electrical potential in this region with these boundary conditions must be spherically symmetric. (The potential must be of the form of the potential of a point charge located at the center of the shell.) This in turn requires the electric field outside the conductor to be spherically symmetric. Since the charge density at the outer surface of the conductor is proportional to the electric field at the surface, the charge density on the outer surface must be uniformly distributed over the surface.

Note that this charge density on the outer surface produces no electric field inside the outer surface. So, it has no influence on how charge is distributed on the inner surface of the shell.

Now consider the inner surface of the shell. The superposition of the electric field of the point charge inside the cavity and the induced charge on the inner surface must add to zero at every point inside the conducting material. If the point charge is not at the center of the shell, then you should be able to argue that the surface charge density on the inner surface cannot be uniformly distributed and still have zero net electric field inside the conducting material.
 
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  • #17
TSny said:
The induced charge will be uniform on the outer surface, but non-uniform on the inner surface.

Hello TSny

Is the potential inside a spherical shell having uniform charge distribution and a shell having non uniform charge distribution same ?

In this question , the charge distributed on the inside surface is not uniformly distributed .

Is the potential inside still given by Kq/R ?
 
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  • #18
Tanya Sharma said:
Is the potential inside a spherical shell having uniform charge distribution and a shell having non uniform charge distribution same ?

In this question , the charge distributed on the inside surface is not uniformly distributed .

Is the potential inside still given by Kq/R ?

Hello, Tanya.
For an arbitrary point inside the cavity, the potential for a charge Q distributed uniformly on the inner surface would not equal the potential at that point if Q is distributed non-uniformly. However, for the point at the center of the shell, you can show that the potential is kQ/R whether or not Q is distributed uniformly. [A bit of charge dQ on the inner surface produces the same potential at the center no matter where on the inner surface dQ is located. Therefore, V at the center will be the same no matter how the bits of charge dQ are distributed.]

So, the answer you gave was correct and nicely stated.
 
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