What are Some Unresolved Questions in Electrostatics?

In summary, according to the author, a point charge at the center of a sphere will produce a uniform field, while charges at other points will produce distorted field lines. The author also explains that when charges are transferred from a charged conductor to a neutral one, the use of induction will create polarization of the charge and then grounding will not work. However, if charges are connected with a conducting wire, only half of the charge will be transferred. Gauss's law states that if a volume charge is enclosed by a uniform field, then the electric field is also uniform inside the body, but this is not always the case. Gauss's law is used to calculate electric potential at the center or corner of a uniformly charged cube, but this is not
  • #1
phymathlover
11
0
I have a few questions pertaining to some concepts in electrostatics, I'd be grateful if someone would help me out.

1) When we place a positive point charge inside a hollow conducting sphere, at its center, field lines emerge uniformly from the sphere. Okay, easy enough. Now, if we place the point charge inside the hollow sphere at a point other than the center, negative charge is induced on the inner surface of the sphere and equal amount of positive charge on its outer surface, now field lines inside the sphere are distorted i.e in a diagram we represent more closely spaced field lines where the charge is closer to the surface of the sphere than at points where it is relatively farther. So far so good. But when the lines emerge from the sphere they emerge out uniformly just as if the charge was at the center. How does that happen? An explanation I found said that the induced negative charge does not produce a field so they emerge uniformly. That just went over my head. I'd appreciate it if someone could explain why to me, explain like you would to a toddler XD

2)How can you transfer complete charge from a charged conductor to a neutral one? (Same sign charge)
Edit - Use of induction to create polarization of charge (charge of the neutral body opposite in sign to that of the charged body collects around the side closer to the charged body and the similar charges get repelled to the other side of the neutral body. So one side of the body has partial positive charge and the other side has equal partial negative charge, so the body still remains electrically neutral) and then earthing it will obviously not work.
But if I connect the bodies with a conducting wire, only half of the charge will be transferred ...
Is there any other way?

3) If I place a conductor in a uniform electric field and then cut out a cavity in the conductor, will the electric field exist in the cavity? I remember reading somewhere that it doesn't but why not? I consulted a lot of books but most of them simply stated this or said that it can be proved by the use of higher mathematics :P

4) Using Gauss law if we get enclosed charge = 0, does it necessarily imply that electric field is also 0 inside the body? And is the field under consideration internal (caused by the charge inside the body) or external (by other charges existing outside the Gaussian surface)? So will external field exist inside the Gaussian surface even if charge enclosed is 0?
Edit - Okay so the electric field under consideration is both internal and external. So q being 0 in Gauss's law implies that the number of field lines entering is equal to the number of field lines leaving the body? And also since according to Gauss law the net charge enclosed is proportional to the dot product of E(vector) and ds(vector) so it merely implies that the angle between the area vector and field vector is 90? So that doesn't mean field doesn't exist? Are my speculations correct?
And also, shouldn't a Gaussian surface necessarily be an equipotential (potential same at all points on the surface) surface? Any supporting or contradictory examples with the explanation would be helpful

5) If you take a uniformly charged (volume charge) sphere and consider a concentric smaller sphere inside the bigger sphere then what kind of field lines emerge from the smaller sphere and why? Will the charge outside the smaller sphere affect the field lines emerging? (I know field lines are imaginary but how will I draw them in this case?

6) How do I go about calculating electric potential at the center or corner of a uniformly charged (volume charge) cube? I mean the field emerging will be non-uniform right? And my friend told me I can't use Gauss's law ... I only know that V=kq/r and rho=q/(side)cube

7) When we place a point charge at the corner of a cube, we take 1/8 part of it to be inside the cube and when we place it on a face, we take 1/2 part to be inside. That is understandable with a sphere but a point charge is a ... point! So how do we start considering it like a sphere and making it into parts?

8) I'd like it if someone could show me what a non-uniform field looks like and what kind of body it emerges from. I couldn't find it in any of my reference books and googling it turned out to be an epic fail

I'd appreciate it if any existing conceptual mistakes in my explanations are pointed and any misconception I may be having cleared.
Thank you to anyone who takes the time to explain :)
PS - I really hope this isn't considered a homework question because I have problems with the concepts. Reason for not mentioning book names - They're books by Indian authors not sold internationally
 
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  • #2
3) The reason the hollow cavity won't have an electric field inside is because positive charge in the conducting material will move toward the negative side of the external field, while negative charge in the conductor will move toward the positive side of the external field. So now the positive charge on the negative side cancels with the negative charge on the positive side...this is kinda hard to explain in words but draw a picture and you should see it more clearly.
 
  • #3
copernicus1 said:
3) The reason the hollow cavity won't have an electric field inside is because positive charge in the conducting material will move toward the negative side of the external field, while negative charge in the conductor will move toward the positive side of the external field. So now the positive charge on the negative side cancels with the negative charge on the positive side...this is kinda hard to explain in words but draw a picture and you should see it more clearly.

I have no idea what the positive and negative sides of external fields are -_-
But anyway, even if the charges cancel each other, the outer field shouldn't be affected by the presence or absence of charges right? I mean sure, the curve of the field lines changes and all but the field still exists ... I mean that should just imply that number of field lines entering the cavity is equal to the number exiting. Ahh ... I'm confused!

I had a brain flash! If we assume that field exists in the cavity, this would imply that there is a potential difference between any two points surrounding the cavity (because field flows from higher potential to lower potential) But the points surrounding the cavity belong to the conductor and we know that a conductor is an equipotential surface i.e potential difference is zero. So this contradicts our initial assumption and hence field can not exist inside a cavity within a conductor.
Is this a proper reason?

Please please do post other possible reasons. I could only come up with this one.
 
  • #4
phymathlover, your post with the questions is excessively long. You will have better chance of getting response if you ask shortly, one or two questions at time.

I will try to answer the first question.

If there are no charges inside the closed metallic cavity, the electric field is zero inside the cavity as well as inside the metal.

But if there are charges inside, the field inside is non-zero, and the field in the metal is still zero. This is due to the Gauss law

$$
\oint \mathbf E \cdot d\mathbf S = Q/\epsilon_0
$$
and the fact the within the metal static electric field is always zero.

Are you familiar with these?


But when the lines emerge from the sphere they emerge out uniformly just as if the charge was at the center. How does that happen? An explanation I found said that the induced negative charge does not produce a field so they emerge uniformly. That just went over my head. I'd appreciate it if someone could explain why to me, explain like you would to a toddler XD

In order to nullify the field in the metal to zero, there has to be charge density on the inner surface that produces additional field of exactly opposite value than the field due to the charges inside. This cancellation occurs for all points outside the inner surface, not just in the metal.

The charges on the outer surface are therefore undisturbed by the inner charges, so they remain uniformly distributed over the sphere and the resulting field is the same as before - radial.
 
  • #5
4) No. Zero enclosed charge does not imply that the electric field is zero everywhere inside the enclosed volume. For example in a uniform electric field, Gauss's law implies no charge inside any closed surface.

No, a Gaussian surface need not be an equipotential surface. You are free to choose whatever surface (that encloses a volume) as your Gaussian surface; you just choose one that is convenient for solving the problem at hand. For instance if there is spherical symmetry then you would (probably) want to choose a sphere as your Gaussian surface.
 
  • #6
7) Uniform field means direction and magnitude of the field are everywhere the same, such as between plates of parallel plate capacitor; anything else is non-uniform, e.g. field from point charge.
 
  • #7
phymathlover said:
I had a brain flash! If we assume that field exists in the cavity, this would imply that there is a potential difference between any two points surrounding the cavity (because field flows from higher potential to lower potential) But the points surrounding the cavity belong to the conductor and we know that a conductor is an equipotential surface i.e potential difference is zero. So this contradicts our initial assumption and hence field can not exist inside a cavity within a conductor.
Is this a proper reason?

You nailed it, phymathlover! That is the correct reasoning (or one of the correct reasonings) to answer your question. Indeed very few books develop this point that far and usually end up restricting themselves to say that external fields cannot enter a conductor's cavity.

Now, just for fun, we could imagine that someone came to you and said that the field lines are so exotic inside the cavity that even the path integrals that do not run on the cavity's walls are zero and, consequently, your previrous argument would not be valid in this case. But then you think for a couple of seconds and reply to that person pointing out that, if that was to be true, it would mean that the the field lines would be such that the flux of the electric field on a closed surface located inside the cavity would be different of zero, implying the existence of some net charge. However, by assumption there is no charge inside the cavity, and therefore the fields could not be that exotic. :D
 
  • #8
Jano L. said:
phymathlover, your post with the questions is excessively long. You will have better chance of getting response if you ask shortly, one or two questions at time.

I will try to answer the first question.

If there are no charges inside the closed metallic cavity, the electric field is zero inside the cavity as well as inside the metal.

But if there are charges inside, the field inside is non-zero, and the field in the metal is still zero. This is due to the Gauss law

$$
\oint \mathbf E \cdot d\mathbf S = Q/\epsilon_0
$$
and the fact the within the metal static electric field is always zero.

Are you familiar with these?




In order to nullify the field in the metal to zero, there has to be charge density on the inner surface that produces additional field of exactly opposite value than the field due to the charges inside. This cancellation occurs for all points outside the inner surface, not just in the metal.

The charges on the outer surface are therefore undisturbed by the inner charges, so they remain uniformly distributed over the sphere and the resulting field is the same as before - radial.

I can now see the ridiculous length of my post XP
Are we allowed to post many threads in the same forum at the same time? Like if I'd opened a separate thread for each one of my questions?

I am having a little trouble visualizing this ... "This cancellation occurs for all points outside the inner surface, not just in the metal" So what does that mean? The surrounding medium also gets ionized and produces a field?
I am also failing to comprehend everything after that because if stuff outside the metal is producing fields, the outer charge should also be disturbed ...
I shall stare at this and try to understand ... lol XD forgive my slowness

tomfy said:
4) No. Zero enclosed charge does not imply that the electric field is zero everywhere inside the enclosed volume. For example in a uniform electric field, Gauss's law implies no charge inside any closed surface.

No, a Gaussian surface need not be an equipotential surface. You are free to choose whatever surface (that encloses a volume) as your Gaussian surface; you just choose one that is convenient for solving the problem at hand. For instance if there is spherical symmetry then you would (probably) want to choose a sphere as your Gaussian surface.

So there are no restrictions imposed on the choosing of Gaussian surfaces? It just has to be a closed surface and we can take any kind of surface which helps us solve our problem?

Zag said:
You nailed it, phymathlover! That is the correct reasoning (or one of the correct reasonings) to answer your question. Indeed very few books develop this point that far and usually end up restricting themselves to say that external fields cannot enter a conductor's cavity.

Now, just for fun, we could imagine that someone came to you and said that the field lines are so exotic inside the cavity that even the path integrals that do not run on the cavity's walls are zero and, consequently, your previrous argument would not be valid in this case. But then you think for a couple of seconds and reply to that person pointing out that, if that was to be true, it would mean that the the field lines would be such that the flux of the electric field on a closed surface located inside the cavity would be different of zero, implying the existence of some net charge. However, by assumption there is no charge inside the cavity, and therefore the fields could not be that exotic. :D

What are exotic field lines? 8o lol XD
Are they like complicated ones or something?
Path integrals? Is that like the path followed by the field lines?

I got the basic idea though, complicated field, flux non-zero therefore enclosed charge non-zero
So disproved!
 
  • #9
phymathlover said:
What are exotic field lines? 8o lol XD
Are they like complicated ones or something?
Path integrals? Is that like the path followed by the field lines?

I got the basic idea though, complicated field, flux non-zero therefore enclosed charge non-zero
So disproved!
Yes, I meant "exotic" as a synonym for "very odd", for example.

A path integral is an integral performed over a mathematical path. You should recognize it on your notes or books since it is exactly the mathematical tool used to calculate the variation of the electrical potential:

[itex]\int\limits_A^B \mathrm{\mathbf{E}}\cdot\mathrm{d}\mathbf{l} = V_{A} - V_{B}[/itex]

That's the reason why your previous argument was correct. If A and B are two distinct points on the cavity's surface, you can perform this integral through many different paths. If you choose a path that is contained in the cavity's surface, then [itex]\mathrm{\mathbf{E}}\cdot\mathrm{d}\mathbf{l} = 0[/itex] at every point of this path, because the Electric Field is always perpendicular to a conductor's surface (be it internal or external). Consequently [itex]V_{A} = V_{B}[/itex], which is to say that the cavity's walls are an equipontetial surface. Now, if you perform that same integral over a path that runs inside the cavity, but still departs from A and arrives at B, because we supposed there was a non-zero field inside the cavity, you would have to conclude that [itex]V_{A} \neq V_{B}[/itex]. But the surface is an equipotential as we've just seen! So you get to the conclusion that it is not possible there are fields inside the cavity! :DWell, I am glad you got the idea. I wish you good luck on your studies! ;)Zag
 
  • #10
Zag said:
Yes, I meant "exotic" as a synonym for "very odd", for example.

A path integral is an integral performed over a mathematical path. You should recognize it on your notes or books since it is exactly the mathematical tool used to calculate the variation of the electrical potential:

[itex]\int\limits_A^B \mathrm{\mathbf{E}}\cdot\mathrm{d}\mathbf{l} = V_{A} - V_{B}[/itex]

That's the reason why your previous argument was correct. If A and B are two distinct points on the cavity's surface, you can perform this integral through many different paths. If you choose a path that is contained in the cavity's surface, then [itex]\mathrm{\mathbf{E}}\cdot\mathrm{d}\mathbf{l} = 0[/itex] at every point of this path, because the Electric Field is always perpendicular to a conductor's surface (be it internal or external). Consequently [itex]V_{A} = V_{B}[/itex], which is to say that the cavity's walls are an equipontetial surface. Now, if you perform that same integral over a path that runs inside the cavity, but still departs from A and arrives at B, because we supposed there was a non-zero field inside the cavity, you would have to conclude that [itex]V_{A} \neq V_{B}[/itex]. But the surface is an equipotential as we've just seen! So you get to the conclusion that it is not possible there are fields inside the cavity! :DWell, I am glad you got the idea. I wish you good luck on your studies! ;)Zag

Oh! Lol XD, now I recognize it XD
I've used a path integral quite a few times myself while solving questions ... I just didn't know it was called that XP
Thank you for clearing that up for me :)

I'll love you for eternity if you could answer the remaining questions too :look:
 

Related to What are Some Unresolved Questions in Electrostatics?

1) What is the difference between conductors and insulators in electrostatics?

In electrostatics, conductors are materials that allow the flow of electric charges, while insulators do not. This is due to the difference in the number of free electrons in each material. Conductors have a larger number of free electrons, making it easier for charges to move through them, while insulators have fewer free electrons, making it difficult for charges to move.

2) How does the distance between two charges affect the electrostatic force between them?

The electrostatic force between two charges is inversely proportional to the square of the distance between them. This means that as the distance between the two charges increases, the force decreases. Similarly, as the distance decreases, the force increases.

3) Can electrostatic forces act through a vacuum?

Yes, electrostatic forces can act through a vacuum. This is because electric charges create an electric field around them, which can exert a force on other charges, even if there is no physical contact between them.

4) What is the difference between electric potential and electric potential energy?

Electric potential is the amount of potential energy per unit charge at a specific point in an electric field. It is measured in volts and represents the work needed to move a unit of charge from one point to another. Electric potential energy, on the other hand, is the energy that a charge possesses due to its position in an electric field.

5) How does the charge distribution affect the electric field in a conductor?

In a conductor, the charges are free to move, and they will distribute themselves in such a way that the electric field inside the conductor is zero. This is because any excess charge will repel other charges and redistribute themselves on the surface of the conductor, creating a net zero electric field inside.

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