How Does Earthing Affect Charge Distribution on Spherical Shells?

[gracy]

From post #9 I am quoting

If shell B has some charge q' on its surface

But why is it needed?There is no charge inside the shell C in the first place i.e without this q' .So no field inside.

 

[gneill]

What I am asking is can that hollow conductor (hollow shell :shell containing cavity inside)be treated as a cavity inside the conductive shell?

[/QUOTE]

A "shell" is a single contiguous layer that encloses some space. It has two surfaces: an inside surface and an outside surface. I'm having trouble distinguishing a shell from your "hollow shell", since they should be identical, with the "hollow" being redundant. Two nested shells can have a hollow space between them. Perhaps that's what you mean?

In your drawing what I'm seeing is an outer shell (gold color) with a cavity, and in that cavity is another, single very thin shell represented by the line of a circle.

A shell can hold a charge, but a void (cavity) cannot. So in general a shell is not the same as a cavity.

 

[gracy]

A shell can hold a charge, but a void (cavity)can not.

If you put a charge in the cavity there will be a field in the cavity due to that charge.

So do you mean a cavity can have field inside it but it (cavity)can not have charge inside it.Because charge will migrate to the surface of conductor.Hence,charges migrate from cavity but field stays.

 

[gracy]

Thanks for your patience @gneill.I am extremely sorry for my rapid questions.

 

[gneill]

From post #9 I am quoting
But why is it needed?There is no charge inside the shell C without this q' .

I can't tell you why it's needed since you haven't presented the rest of the context in which that diagram was found. For all I know they go on to prove that q' must have some particular value for the situation shown.

So do you mean a cavity can have field inside it but it (cavity)can not have not charge inside it.Because charge will migrate to the surface of conductor.Hence,charges migrate from cavity but field stays.

A cavity is empty space. Empty space can support fields. You can place charges (particles) inside a cavity and their fields will fill the cavity. But the cavity itself (empty space) cannot hold a charge. The best you can do is place things with charges in it, or try to place a charge on its boundary surface. But that boundary surface is part of a conductive shell, so they would be free to move away.

 

[gracy]

I can't tell you why it's needed since you haven't presented the rest of the context in which that diagram found.

Only this much information has been given.Figure (2) and charge distribution has been given as a hint.

 

[Dale]

Closed pending moderation.

 

[gneill]

Thread re-opened.

 

[ehild]

Closed pending moderation.

I wonder why?

 

[gneill]

I wonder why?

There was some concern that the thread was being spread into other threads with the conversations and results crossing each other. This makes following anyone thread a tad confusing, and moderating the ensemble somewhat problematic. Merging the threads was discussed, but it was decided that it would not give a satisfactory or more elegant result. So for now the thread is re-opened and any new fragmentation will be discouraged.

 

[ehild]

Only this much information has been given.Figure (2) and charge distribution has been given as a hint.

.
From the OP:

Figure (1)shows three concentric thin spherical shells A,B and C of radii a,b and c respectively.

The shells are thin, but still they have inner and outer surfaces. The shell makes a wall that surrounds the enclosed cavity. There is no charge, and the electric field is zero inside the wall, but there can be surface charge on both surfaces. The following is better for Fig. 2 as it was in the OP as it indicates that the charges belong to the surfaces.

Shell B was not given any charge, but it is grounded, the shell can have nonzero net charge.

 

[ehild]

There was some concern that the thread was being spread into other threads with the conversations and results crossing each other. This makes following anyone thread a tad confusing, and moderating the ensemble somewhat problematic. Merging the threads was discussed, but it was decided that it would not give a satisfactory or more elegant result. So for now the thread is re-opened and any new fragmentation will be discouraged.

Thank you gneill, for the explanation, and also for re-opening the thread.

 

[gracy]

the shell can have nonzero net charge.

But it is not compulsory that every earthed shell would have some non zero net charge.Why surface of shell B contain charges,what if the charges are not there?

 

[ehild]

But it is not compulsory that every earthed shell would have some non zero net charge.Why surface of shell B contain charges,what if the charges are not there?

A single earthed shell is not charged, but B encloses a charged shell, and is surrounded by an other charged shell. If you derive the potential of all shells, you get how much charge B must have to be at zero potential.

 

[gracy]

If you derive the potential of all shells, you get how much charge B must have to be at zero potential.

But I was not told to find potentials so I did not consider it ,I believe that's why I got stuck.
Does it mean whenever I will find such problem when there is not single earthed shell ,I'll assume some charge on it's surface later with the help of potentials ,I will find how much charge B must have to be at zero potential?
Right?

 

[gracy]

What if the problem does not mention "thin",solution would be same?

 

[ehild]

But I was not told to find potentials so I did not consider it ,I believe that's why I got stuck.

The problem asked you

Find the charge appearing on the surfaces of B and C.

, which means you need q' in terms of q. You should have figured out how to do it. You have to know, how to get the electric field of charged shells and potential difference between them, and how to get the potential of the outer shell C. Deriving the expresion for the potentials of the shells, you get an equation for q' in terms of q.

Does it mean whenever I will find such problem when there is not single earthed shell ,I'll assume some charge on it's surface later with the help of potential ,I will find how much charge B must have to be at zero potential?
Right?

Right. If a charged body is near an earthed conductor, the conductor gets charge from the earth.

 

[ehild]

What if the problem does not mention "thin",solution would be same?

Thin means the thickness of the shell is small with respect to the radius, so we can speak about "the radius of the shell". If they were not thin, the inner and outer surfaces would have different radii. Those radii should be given, and the solution would be different.

 

[gracy]

the inner and outer surfaces would have different radii.

Will it be a problem?

 

[ehild]

No problem, but you need to take them into account when calculating the potential of the shell.

 

[gracy]

r.Hence,charges migrate from cavity but field stays.

But according to the following video

charge +q stays in the metal cavity .

Does it mean whenever I will find such problem when there is not single earthed shell ,I'll assume some charge on it's surface later with the help of potentials ,I will find how much charge B must have to be at zero potential?

How do we know that outer surface of B would have positiv charge i.ewhy not we assumed -q' instead of q'?

 

[cnh1995]

How do we know that outer surface of B would have positive charge i.e.why not we assumed -q' instead of q'?

I believe it is because the net field inside the thickness of shell B must be 0 since it is a metal.

 

[Dale]

How do we know that outer surface of B would have positiv charge i.ewhy not we assumed -q' instead of q'

The video itself explains why very clearly. What does the video say?

 

[gracy]

How do we know that outer surface of B would have positiv charge i.ewhy not we assumed -q' instead of q'?

It is not from the video.It is fro my OP.

 

[Dale]

Then why did you bring in the video at all? Honestly gracy, that is irritating.

Thread closed. Please spend more time on your own before posting and then put some additional effort into making clear and focused threads.