i wanted to ask why the electric field inside a hollow conductor zero throughout and not just at the centre.
Thread moved to homework help. zarbanx, please tell us what you know about electric fields in the presence of conductors. What happens to the electrical charges in conductors when an electric field is present? We do not do your studying for you -- you must show some work and effort before we can offer tutorial assistance.
okk as u say well i have done a lot of work and research i know tht there is no electric field inside a conductor bt i am not able to prove it mathematically and moreover electrical charges in conductors move to the surface becoz no electric field is there in a conductor becoz if there is a field then charges will move to neutralizze it.when an external electrical field is present then charges rearrange tso that no electric field is there in the conductor bt still mathematically i am not able to prove it
Well, if you open up your textbook, and skim to the sections on Gauss' Law and spherical distributions of charge, do you see some hints about what is going on with a hollow conductor? What kind of boundary condition volume is typically used with Gauss' Law? Since electric charges are free to move in a conductor in response to an Electric field, what will the charge distribution look like on a hollow conductor when embedded in an electric field? That should be enough hints to get you going. If your textbook isn't showing this to you for some reason (which text are you using?), try searching some at wikipedia.org
Well H&R definitely talks about this. Remember, we do not do your work for you. Post your work, and we can offer tutorial help.
The field is zero inside only if any charge is evenly distributed on the surface. That's a mathematical theorem, sorry I don't have the proof handy. But when you measure the electric field inside a charged sphere, the charge you use might be large enough to redistribute the surface charge. In this case the electric field will not be zero. Only if you measure at the centre. Now, what happens if there is an external field that is causing a redistribution of surface charges, I don't know. It may be another just as unexpected theorem. If you find proof for this, let me know.
There is an analogy to this that you might find helpful; it has to do with the gravity force acting on a person inside a hollowed-out shell of a planet. By symmetry the force must be zero when a person is at the center, but it is not so intuitive to see that the force is zero everywhere inside the shell. Isaac Newton used what is called "Shell Theorem" to rigorously prove some important things about spherical shells, one of which is what I mention above, and another of which is that any spherical object can be modeled as a point mass when you are located outside the object. So go to Wikipedia as berkeman suggested, and read about Shell Theorem. You will find more than enough mathematical rigor, believe me. And since gravity and electrostatic force are both inverse-square forces, the methods are almost the same for analyzing a conducting spherical shell. From there move on to non-spherical shells, if you like.
okk thanx i was thinking tht electric field cease to exist inside the shell bt now i know tht they mutually cancel out.........right
prob solved bt ulysses said tht charge's uniform distribution is necessary for electric field to be zero inside the sphere .....................is tht necessary?
Yes, Shell Theorem relies explicitly on a uniform distribution of mass/charge/whatever. If a thin spherical plastic shell had a small section made of lead, for example, that section would clearly exert a stronger force on a person inside and ruin the symmetry.
Merryjman, are you familiar with the math involved in here? I have got stuck in another similar problem: https://www.physicsforums.com/showthread.php?t=212711
I'm not sure that's true. What happens in an external field is that the conductor will become polarized, and it polarizes in such a way that the field inside is still zero.
The thing is, proof for this statement still eludes us. It is certainly quite unexpected, as unexpected as the Shell Theorem, so I'd appreciate some proof.
I wrote some of the Wikipedia article on Shell Theorem, so I am familiar with most of the math. I'll look at that thread. As far as the conductor goes, I'd have to look in Griffiths to find a rigorous mathematical proof (I seem to remember LeGendre polynomials being a part of this), but I can at least hand-wave it. Since electrons on a perfect conductor are completely free to move, they are going to rearrange themselves until they are in equilibrium. The presence of an external field is going to make electrons move upfield because they are being attracted to (or repelled from) whatever charge distribution set up the external field. That's just a consequence of Coulomb's Law, which can't be proved - this so-called "Law" is based on experimental observation of the force between charged particles, not on any sort of mathematical construct. Newton's "Law" of gravitation is the same - it was based on observational data, and more modern data have shown that NULoG is incomplete; hence general relativity. If the electric field inside a conductor was NOT zero, then there would be a force acting on the mobile charges, and so they would rearrange until the force WAS zero.
It's conceivable the total force is zero on the surface, where each infinitesimal charge sits, and non-zero inside.
But if the force was non-zero inside, charges would still be moving, and the situation would not be electrostatic. Electric fields are nonzero in current-carrying wires, for example. If you were looking at the conductor at the instant the external electric field was applied, there would be internal fields and currents as the charges rearranged.
Why? The field inside need not be identical to the field on the surface. Might be zero inside and non-zero on the surface or vice versa when equilibrium is reached.