Applying Gauss's Law Several Times

In summary, the conversation discusses the concept of electrostatic equilibrium and the use of Gaussian surfaces to calculate electric fields. It is explained that when using a Gaussian surface inside a conducting, hollow sphere, the electric field is zero due to the cancellation of charges on the inner surface. The insulator within the sphere can induce this equilibrium and the amount of induction can be quantified by the amount of charge on the inner and outer surfaces of the conductor.
  • #1
Bashyboy
1,421
5

Homework Statement


I am working on a problem that is similar to this one:

https://www.physicsforums.com/showthread.php?t=598914

Homework Equations


The Attempt at a Solution


When r < a, why doesn't the electric fields generated by the concentric conducting, hollow sphere affect the Gaussian surface at that point??

Why is the electric field at a point in the concentric conducting, hollow sphere zero? After all, it doesn't specify that it is in electrostatic equilibrium.

Also, what is the difference between the entire system being in electrostatic equilibrium and just the concentric conducting, hollow sphere being in electrostatic equilibrium? Can an insulator be in electrostatic equilibrium?
 
Physics news on Phys.org
  • #2
Bashyboy said:

Homework Statement


I am working on a problem that is similar to this one:

https://www.physicsforums.com/showthread.php?t=598914



Homework Equations





The Attempt at a Solution


When r < a, why doesn't the electric fields generated by the concentric conducting, hollow sphere affect the Gaussian surface at that point??

Gauss says consider the charge INSIDE the Gaussian surface only. What is the charge inside a spherical surface for r < a?



Why is the electric field at a point in the concentric conducting, hollow sphere zero? After all, it doesn't specify that it is in electrostatic equilibrium.
It's not zero in the hollow area. It is in the conducting area. And you are to assume equilibrium.

Also, what is the difference between the entire system being in electrostatic equilibrium and just the concentric conducting, hollow sphere being in electrostatic equilibrium? Can an insulator be in electrostatic equilibrium?
You will probably never have to worry about electrostatic disequilibrium. You were given that the insulating sphere had a uniform charge density. Therefore it's in equilibrium.
 
  • #3
So, you are saying that if I was to generate a Gaussian surface inside of the conductor (say, a sphere), then the flux through this surface is zero, because the electric field through this surface is zero? How can this be? I know that the charge isn't distributed inside of the conductor, the charge resides on the surface, but the Gaussian surface would still enclose the insulator, which has a charge.

EDIT: Something just came to my mind: does the insulator induce the electro-static equilibrium? And if the insulator had a positive charge, it would cause the electrons in the conductor to move to the surface closest to the insulator? Leaving the other surface "less" negative?
 
  • #4
Bashyboy said:
So, you are saying that if I was to generate a Gaussian surface inside of the conductor (say, a sphere), then the flux through this surface is zero, because the electric field through this surface is zero?
Yes.

How can this be? I know that the charge isn't distributed inside of the conductor, the charge resides on the surface, but the Gaussian surface would still enclose the insulator, which has a charge.
Charge can rearrange itself on the inner surface of the conducting shell to cancel any charge placed within the hollow of the sphere.

EDIT: Something just came to my mind: does the insulator induce the electro-static equilibrium? And if the insulator had a positive charge, it would cause the electrons in the conductor to move to the surface closest to the insulator? Leaving the other surface "less" negative?
Yes. Any charge placed inside the conducting sphere will induce a charge on the inner surface that cancels its field within the conductor (and for all points beyond).
 
  • #5
So, using the diagram given in the link, if I was to place a Gaussian sphere directly between the insulator and conductor, the electric field lines emanating from the insulator will "hit" the sphere from the inside, and the electric field lines emanating from the conductor "strike" the very surface of the Gaussian sphere; and so, the electric fields combine to cancel out?

And is this the reason why we don't have to consider the electric field of the conductor when generating a Gaussian surface that meets the condition of r < a, because the electric fields lines never even make it that far?

EDIT: Also, is there a way of quantifying the amount of induction that the insulator affects? Would it just be equal and opposite to the charge of the insulator? And the remaining charge would reside on the outside?
 
Last edited:
  • #6
For example, say the insulator has a charge of Q1 = 3.60 C, and the conductor had a total charge of -2.00 C. Would the inner surface of the conductor would have a charge of Q2 = -Q1 ---> Q2 = -3.60 C; and would this then imply that Q3, the outer surface of the conductor, would have a charge of Q2 + Q3 = -2.00 C ----> Q3 = 1.6 C?
 
  • #7
Bashyboy said:
For example, say the insulator has a charge of Q1 = 3.60 C, and the conductor had a total charge of -2.00 C. Would the inner surface of the conductor would have a charge of Q2 = -Q1 ---> Q2 = -3.60 C; and would this then imply that Q3, the outer surface of the conductor, would have a charge of Q2 + Q3 = -2.00 C ----> Q3 = 1.6 C?
That is correct.
 
  • #8
Why is it correct, if you don't mind me asking?
 
  • #9
Is there some law or principle of induction?

Bashyboy said:
So, using the diagram given in the link, if I was to place a Gaussian sphere directly between the insulator and conductor, the electric field lines emanating from the insulator will "hit" the sphere from the inside, and the electric field lines emanating from the conductor "strike" the very surface of the Gaussian sphere; and so, the electric fields combine to cancel out?

And is this the reason why we don't have to consider the electric field of the conductor when generating a Gaussian surface that meets the condition of r < a, because the electric fields lines never even make it that far?

EDIT: Also, is there a way of quantifying the amount of induction that the insulator affects? Would it just be equal and opposite to the charge of the insulator? And the remaining charge would reside on the outside?

Is it possible for someone to answer these questions I have from an earlier post?
 
Last edited:
  • #10
Bashyboy said:
For example, say the insulator has a charge of Q1 = 3.60 C, and the conductor had a total charge of -2.00 C. Would the inner surface of the conductor [STRIKE]would [/STRIKE] have a charge of Q2 = -Q1 ---> Q2 = -3.60 C; and would this then imply that Q3, the outer surface of the conductor, would have a charge of Q2 + Q3 = -2.00 C ----> Q3 = 1.6 C?

Doc Al said:
That is correct.

Bashyboy said:
Why is it correct, if you don't mind me asking?
The reason that we know the charge on the inner surface has equal magnitude and opposite sign of the charge on the inner charged sphere, is that if we use a Gaussian surface that lies within the conducting material, such as a sphere with radius between b & c, the E field is zero everywhere on the Gaussian surface so the net charge within the Gaussian surface is zero.
 
  • #11
So, would electric field at a radius of a < r < b be zero?

EDIT: And would it be zero because the number of electric field lines radiating out of the Gaussian surface equals the number of electric field lines coming in?
 
Last edited:
  • #12
Bashyboy said:
So, would electric field at a radius of a < r < b be zero?

EDIT: And would it be zero because the number of electric field lines radiating out of the Gaussian surface equals the number of electric field lines coming in?

No. Put a Gaussian spherical surface anywhere on a < r < b and what is the charge inside that surface? And so what does Gauss say about the E flux thru that surface?
 
  • #13
Bashyboy said:
So, would electric field at a radius of a < r < b be zero?
No. The field there would equal that from the charged insulator. The induced charge on the inner surface of the conducing shell will have no field for r < b.
 
  • #14
So, the flux through a Gaussian only depends on the charge enclosed, further implying that it only depends on the electric field emanating from that enclosed charge?
 
  • #15
Bashyboy said:
So, would electric field at a radius of a < r < b be zero?

EDIT: And would it be zero because the number of electric field lines radiating out of the Gaussian surface equals the number of electric field lines coming in?
Whoa !

E field lines originate from Positive and terminate on Negative charge. The negative charge on the inner surface of the conductor does not interfere with the electric field at a < r < b .

So the field at a radius of a < r < b is not zero. Gauss's Law says the flux though a sphere of such radius is not zero.
 
  • #16
Doc Al said:
No. The field there would equal that from the charged insulator. The induced charge on the inner surface of the conducing shell will have no field for r < b.

Why won't it have a field there?
 
  • #17
Bashyboy said:
So, the flux through a Gaussian only depends on the charge enclosed, further implying that it only depends on the electric field emanating from that enclosed charge?
The total flux calculated over the entire closed Gaussian surface depends only on the net charge enclosed within that surface. (That's what Gauss's law says.) Note that the field at any given point on the surface may well depend on other charges.
 
  • #18
Bashyboy said:
Why won't it have a field there?
The field produced by a uniform shell of charge is zero at all points within the shell. A result derivable from calculus.
 
  • #19
Bashyboy said:
Why is it correct, if you don't mind me asking?

Just put Gaussian spherical surfaces around a < r < b, at b < r < c, and at r > c.
Inside a < r < b you have charge Q due to the inner sphere. Inside b < r < c you have zero total charge inside that surface since the E field all over that surface has to be zero. Look at my equation below. Inside r > c you have nothing but Q since the conductor is uncharged.

In all cases, total free charge inside a surface = 4πr^2 ε0E.
 
  • #20
I'm sorry, I am still having difficulty with this. If I wanted to calculate the electric field at a point in between the insulated sphere and the conducting sphere--that is, I create a Gaussian sphere around the insulated sphere (a < r < b)--I would have the consider the electric field contribution of both the conducting sphere AND the insulated sphere, or would I just figure out the electric field at that point due to the enclosed charge in the Gaussian sphere, which would be the insulated sphere?
 
  • #21
Bashyboy said:
If I wanted to calculate the electric field at a point in between the insulated sphere and the conducting sphere--that is, I create a Gaussian sphere around the insulated sphere (a < r < b)--I would have the consider the electric field contribution of both the conducting sphere AND the insulated sphere, or would I just figure out the electric field at that point due to the enclosed charge in the Gaussian sphere, which would be the insulated sphere?
Because of symmetry, all you need is Gauss's law. That will tell you that the field is only due to the enclosed charge.
 
  • #22
By symmetry of what? the conductive sphere? How does that make the electric field at that point, due to the conductive sphere, zero?
 
  • #23
Bashyboy said:
By symmetry of what? the conductive sphere? How does that make the electric field at that point, due to the conductive sphere, zero?
Since everything in this problem is spherically symmetric, you can say that the field depends only on the distance from the center. Thus, for any spherical Gaussian surface, such as one drawn between a < r < b, you can treat the field as uniform. Thus you can apply Gauss's law to calculate the field. And that only depends on the charge within the surface, thus you can conclude that the contribution of charges external to the surface must be zero.
 
  • #24
Okay, I think I am beginning to understand. I believe I have just one more question, though. If we were to calculate the electric field for r > c, would we only have to consider the charge of the outer surface of the conductive sphere, because the net charge of the insulated sphere and inner surface of the conductive sphere is zero?
 
Last edited:
  • #25
Bashyboy said:
Okay, I think I am beginning to understand. I believe I have just one more question, though. If we were to calculate the electric field for r > c, would we only have to consider the charge of the outer surface of the conductive sphere, because the net charge of the insulated sphere and inner surface of the conductive sphere is zero?
That is correct. (You of course consider all the charge within your Gaussian surface, which ends up being equal to just the charge on the outer surface of the conductor.)
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in physics that relates the electric field to the charge distribution in a given space. It states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

2. How can Gauss's Law be applied several times?

To apply Gauss's Law several times, you can divide a given space into smaller, simpler shapes where the electric field and charge distribution are known. Then, you can use Gauss's Law for each smaller shape and add up the individual electric fluxes to find the total electric flux through the entire closed surface.

3. What are the benefits of applying Gauss's Law several times?

Applying Gauss's Law several times can simplify complex problems and make them easier to solve. It also allows for the calculation of electric flux through irregularly shaped surfaces that cannot be solved using other methods.

4. Are there any limitations to applying Gauss's Law several times?

Yes, there are limitations to using Gauss's Law multiple times. This method is only applicable to situations with a high degree of symmetry, and it assumes that the electric field and charge distribution are constant over the entire surface being considered.

5. Can Gauss's Law be applied to any type of charge distribution?

Yes, Gauss's Law can be applied to any type of charge distribution, including point charges, line charges, and surface charges. However, the shape and symmetry of the charge distribution will determine the complexity of the calculations needed to apply the law several times.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
325
  • Introductory Physics Homework Help
Replies
26
Views
473
  • Introductory Physics Homework Help
Replies
12
Views
4K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
637
  • Introductory Physics Homework Help
Replies
6
Views
768
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top