2
$\begingroup$

Basically something very similar to these pictures enter image description here

These are from two separate books. The bottom picture says that all the excess charge in an object can be transferred to an already charged metal she'll if the object is touched to the inside of the shell, which I doubtfully assume to mean the same thing as that there is no electric field inside a shell.

The top picture is what explicitly explains a reasoning why there is no field inside a shell. But the problem is I don't get the whole explanation. Basically, it says that

a positive point charge (can be negative as well) is placed inside a good hollow conductor, which is isolated. [This is what confuses me the most-->]Because there can be no field within the metal, the lines leaving the positive charge must end on negative charges on the inner surface of the metal. Thus, an equal amount of of negative charges is induced on the inner surface of the spherical shell. Then since the shell is neutral, a positive charge, +Q, of the same magnitude must exist on the outer surface of the shell.

Ok, there is something veeeerrry contradictory here. If the whole point of this part is to explain why there is no inside the hollow shell, why is it explicitly stated that "the lines leaving...end on negative charges"?

An more irrelevant question is, does distance matter how much charge is induced?

$\endgroup$
  • $\begingroup$ There may be a misunderstanding here. There is no electric field inside a conducting shell IF there is no charge inside. If there is a charge inside, then there is an electric field. Whatever charge is inside or on the outside of the shell will also show as a total charge (i.e. you can't screen total charge), which is what the first image is trying to show. $\endgroup$ – CuriousOne Mar 7 '16 at 1:05
  • $\begingroup$ So if there is already some field in the shell beforehand, not all the field lines from that point charge would all end on the surface? $\endgroup$ – most venerable sir Mar 7 '16 at 1:18
  • $\begingroup$ And it wouldn't look as if " there is no shell at all? $\endgroup$ – most venerable sir Mar 7 '16 at 1:19
  • $\begingroup$ The shell will shield the internal structure of the charges, so you couldn't tell how they are arranged, but it will act as if all the charges are sitting on its surface. $\endgroup$ – CuriousOne Mar 7 '16 at 1:22
  • $\begingroup$ You didn't answer my first question $\endgroup$ – most venerable sir Mar 7 '16 at 1:24
1
$\begingroup$

"Because there can be no field within the metal, the lines leaving the positive charge must end on negative charges on the inner surface of the metal."

enter image description here

The external electric field from the charged object induces charges on the surface of the conductor.
Those induced charges produce an equal in magnitude but opposite in direction electric field resulting in there being no net electric field within the conductor.
This is very much post dates Faraday's interpretation of the ice pail experiment as in Faraday's paper I do not think that there is any mention of the electric field (force) inside the conductors.
What Faraday did show with his ice pail experiment was that the magnitude of the charge causing the induction is equal to the magnitude of the induced charge and that the induced charges resided on the ice pail.

You might find it interesting to read Faraday’s original paper which he wrote over 170 years ago?
The language is old fashioned but the ideas are all there.

$\endgroup$
  • $\begingroup$ It's counterintuitive. The distance doesn't matter? Because it's a sphere? $\endgroup$ – most venerable sir Mar 8 '16 at 21:09
  • $\begingroup$ I mean say the point charge inside the shell is the same size. But the radius of the shell is dramatically increased. Would the induced charge=inducing change? $\endgroup$ – most venerable sir Mar 8 '16 at 21:10
  • $\begingroup$ It is perhaps easier to think of it in terms of electric field lines. If all the electric field lines from the inducing charge alight at the conduction surface then the inducing charge will equal the induced charge.. $\endgroup$ – Farcher Mar 8 '16 at 21:16
1
$\begingroup$

There are some different regions and you need to assign them all names and keep them straight or you'll get them confused.

There is the outside. If you had a solid conductor without a cavity it would be the part that isn't the conductor.

There is the cavity. It's like a hollow region inside the conductor, surrounded by the conductor. It's like the inside of your house if you closed all the doors and windows. It's not the house itself, it's not the walls, it's not the floors or ceiling or roof. It's the region inside with all the air and people. If you replaced the entire outside with a conductor it's the part that now isn't a conductor.

Then the third region is the conductor itself. It's the region that isn't the outside and also isn't the cavity. It's a conducting metal. It has zero electric field.

What about the other regions, do they have zero electric fields. They don't have to, and often they do not.

The field lines come out of the charge in the cavity, go through the region of the cavity and end on the inside surface so they only exist inside the cavity, not in the conductor. Recall those are two different regions.

Next if there is charge on the outside surface the field lines come out of those charges and head out into the outside region. So they exist only on the outside, not in the conductor. Recall that those are also two different regions.

That's it.

But isn't outside just the outer surface of the conductor? Why make it so complicated? The book just says that all the net charges will stay right on the surface.

It's not overly complicated to have different names for different things so you can tell when people are talking about different things.

There are two regions of not-conductor and one is called the outside and one is called the cavity. The surface between the outside and the conductor is called the outside surface. The surface between the cavity and the conductor is called the inside surface. Both surfaces are on the boundary of the conductor. Neither is "inside" the conductor in the sense of being anywhere other than the boundary between the conductor and the not-conductor. You simply need to learn the terminology to learn what people are saying.

Your post clearly showed you had trouble distinguishing when people were talking about the cavity, when they were talking about the outside, and when they were talking about the conducting shell. There is charge on the inside surface (between the cavity and the conductor) and the field lines in the cavity end on the inside surface. They don't go into the conductor. You claimed there was a contradiction, but it's just you misunderstanding when they are talking about each of these three regions. Only one region (the conducting shell) has zero electric field.

$\endgroup$
  • $\begingroup$ But isn't outside just the outer surface of the conductor? Why make it so complicated? The book just says that all the net charges will stay right on the surface. $\endgroup$ – most venerable sir Mar 8 '16 at 21:07
  • $\begingroup$ @Doeser There are two regions of not-conductor and one is called the outside and one is called the cavity. The surface between the outside and the conductor is called the outside surface. The surface between the cavity and the conductor is called the inside surface. Both surfaces are on the boundary of the conductor. Neither is "inside" the conductor in the sense of being anywhere other than the boundary between the conductor and the not-conductor. You simply need to learn the terminology to learn what people are saying. $\endgroup$ – Timaeus Mar 8 '16 at 21:21
  • $\begingroup$ @Doeser Your post clearly showed you had trouble distinguishing when people were talking about the cavity, when they were talking about the outside, and when they were talking about the conducting shell. There is charge on the inside surface (between the cavity and the conductor) and the field lines in the cavity end on the inside surface. They don't go into the conductor. You claimed there was a contradiction, but it's just you misunderstanding when they are talking about each of these three regions. Only one region (the conducting shell) has zero electric field. $\endgroup$ – Timaeus Mar 8 '16 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.