We know that the electric field lines of a unit (point) charge is represented by drawing lines in all the directions radially outwards or inwards (depending on whether the charge is positive or negative) as shown in the figure (1.1).

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Now in a conductor (let us consider a spherical shell) where the charges reside on top of, why isn't there electic field inside the shell if the electic field of charges extend in all directions? If it extends in all directions means that there is some part of the field lines going through the conductor and into the shell.


1 Answer 1


On any point inside the conductor, field line from every charge meet, each with a different direction. It's a simple matter of them cancelling each other out.

Proving that their sum is zero is tedious, but there's a workaround.

In your question, the charges were located at the surface, so you're speaking about a perfect conductor with infinite conductivity. Local Ohm law $\vec{\jmath}=\gamma\vec{E}$ implies that, if $\gamma$ is infinite, $\vec{E}$ has to be zero.

For a non-perfect conductor, a similar result can be obtained if the conductor is in static equilibrium: charges are at rest inside, so the Lorentz force must be zero, so $\vec{E}$ is zero.

  • $\begingroup$ So, if I were to consider a cubical surface (or any other irregular surface that is not spherical) can we say that the field cancels out each other? Beacuse I think the vectors would be irregulary directed and if they aren't cancelled out, there might be some electrical field in it... $\endgroup$ Commented May 14, 2022 at 12:28
  • $\begingroup$ Yes. That's why the computation is usually done indirectly. Remember, however, that your question was about a very special case (conductor with infinite conductivity). With a "real" conductor, skin effect happens, so the electric field is non-zero. $\endgroup$
    – Miyase
    Commented May 14, 2022 at 15:15
  • $\begingroup$ Thank you so much :) $\endgroup$ Commented May 15, 2022 at 16:39
  • $\begingroup$ @SuhasBharadwaj If you consider the answer above as appropriate, please mark it as accepted. It'll help future readers see more quickly that the post was fruitful. Thanks. $\endgroup$
    – Miyase
    Commented May 16, 2022 at 2:31
  • $\begingroup$ Yep, marked it! Thank you for your time. :) $\endgroup$ Commented May 17, 2022 at 11:35

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