If I place a point charge $q$ inside a conductor, The electric field at any point inside it will be non zero ($Kq/x^{2}$). If we draw a Gaussian surface inside the conductor, the net enclosed charge will be $q$ that will provide an outgoing flux. Then why do we say that the electric field inside a conductor is always 0?
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1$\begingroup$ Do you mean a hollow conductor with charge placed inside the cavity, or a bulk conductor? $\endgroup$– aneet kumarCommented Jul 20, 2020 at 7:48
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1$\begingroup$ The enclosed charge will be zero. $\endgroup$– my2ctsCommented Jul 20, 2020 at 7:59
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$\begingroup$ Hollow conductor with charge placed inside it $\endgroup$– Happy UnicornCommented Jul 20, 2020 at 12:06
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$\begingroup$ The point charge q will induce -q charge on the inner surface of hollow cavity and +q charge on the outer surface of conductor. E=0 inside the conductor, the reason being that it has infinite conductivity. $\endgroup$– KP99Commented Jul 18, 2021 at 9:22
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4 Answers
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The free charges will reposition in such a way that the field vanishes, because in a conductor there is no other force acting on them. This also means that any point charge inside a closed surface will be compensated by conducting electrons.
All of this only holds at scale much larger than atomic . At atomic scale the electric field is not zero.
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The field inside the body of the conductor will still be zero. We assume a perfect conductor to have an endless supply of free charges, so regardless of the amount of charge you put inside the hollow of a conductor or outside it, the charges in the body of the conductor will always orient themselves in such a way that the overall field is nullified.
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See Wikipedia's entry on surface charge. If it's truly a perfect conductor, any charge won't be content to stay in the interior. Either is opposite to some charge already present and cancels it out, or else the other charges of the same type will push it inexorably out of the interior of the object to a hypothetically two-dimensional surface layer.
Now, you might think, what if you just keep some charge in the interior by continually supplying more. But we're supposing a perfect conductor that will pass an infinite current to maintain constant voltage. The infinite magnetic field from such a current will prevent any known material from being a superconductor, abolishing the premise. To put it another way, what kind of wire would you use to bring charge into the core of a perfect conductor? Or if you had a battery inside the conductor, how would you separate the + and - terminals?
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The answers so far seem to miss the point by
- assuming the conductor is solid(?) i.e. the charge q becomes part of the conductor (and so is free to move/redistribute itself), and
- do not address the argument in the question: a Gaussian surface which is inside the hollow yet surrounds q will enclose a net charge, and so must have a flux through it.
If that were not the case i.e. surrounding a charge with a conducting shell 'nullified' the field produced by said charge, no matter how large the shell, then
- one could not perform any electrostatic experiments inside a bird cage, or a car, a portable classroom (shipping container), a ship, or in most modern buildings, constructed of reinforced concrete.
- refueling a plane inside a metal hangar would never pose any danger, so there would be no need to connect the plane to the fuel truck.
- there would be no need to wear an earthing strap when pulling a CPU out of its socket while the motherboard was still in the desktop's case.
etc...
Two opposite charges attract each other. Place an alu-foil dome over them, and what: the interaction ceases?
