Assume that we have a charged conductor of any shape but it has an opening so that we can get inside the conductor through that opening .

Will the electric field be zero inside that conductor ?

I am assuming a hollow conductor with an opening.


In general - no, not exactly. The classic rule about the electric field inside a hollow in a conductor arises from the argument that if there is any electric field present on the boundary, charges will redistribute themselves until the entire inside surface is at the same potential. This assumption is violated if there are any regions where the charges are not free to redistribute themselves to where they are needed.

However, this can be "approximately true" if the 'access hole' is small relative to the wavelength of light you are investigating -- e.g. a Faraday Cage made of chicken wire. In this case, radio frequency light inside the cage will be heavily attenuated while visible light passes through more or less unhindered.

| cite | improve this answer | |
  • $\begingroup$ This means if i have a charged conducting cylindrical shell , the E-field will not be zero inside ? I am asking this question because it is a famous problem that we consider the E-field inside the hollow part zero . $\endgroup$ – user207332 Jul 11 '19 at 2:38
  • 1
    $\begingroup$ If the cylinder is infinite, it is still essentially a closed volume on the inside and the result still holds. If the cylinder is impolite enough to be finite, the field inside is decidedly not zero. $\endgroup$ – catalogue_number Jul 11 '19 at 12:34
  • $\begingroup$ I understood it thanks $\endgroup$ – user207332 Jul 11 '19 at 13:19

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.