3
$\begingroup$

Why is the electric field is always zero inside the conducting shell, no matter whether it is charged, grounded, isolated or under external field?

Arguments using only Gauss theorem seems incorrect, because zero net flux does not imply zero electric field. Is uniqueness theorem necessary for this argument?

$\endgroup$
  • $\begingroup$ Yes, uniqueness theorem is essential for showing this. $\endgroup$ – John McVirgooo Mar 7 '17 at 2:54
1
$\begingroup$

1 Electrostatics
$\Rightarrow$ charges not moving.

2 A conductor
$\Rightarrow$ contains a lot of mobile charges.

3 Mobile charge in electric field
$\Rightarrow$ force on mobile charge
$\Rightarrow$ mobile charge accelerates which is contrary to statement 1

So when dealing with electrostatics there is no electric field inside a conductor.

However if there are charges inside a volume which is enclosed by a conducting shell then there will be an electric field inside the volume enclosed by the conducting shell but no electric field inside the actual conducting shell.

$\endgroup$
1
$\begingroup$

According to one source

Electric field is defined as the electric force per unit charge.

So another way to state your question is, why does a unit of charge (electron, proton, etc.) feel no force inside a Faraday cage?

In fact, it does feel a force from every unit of surface of the cage, according to the inverse-square law. Within a given solid angle, every piece of the surface within that angle exerts a force, but if the subject charge is closer to the surface, while the attraction per unit of area is larger, the area itself is smaller by the same factor, so distance from the surface makes no difference. Then if you expand the solid angle to include the whole sphere, those forces all cancel out, so the charge feels no force. The same argument explains why the shape of the cage makes no difference.

Gravity works the same way. Put a mass inside a massive shell, and it feels no gravity, because it feels gravity in all directions, and those cancel out.

$\endgroup$
0
$\begingroup$

The electric field is set up by the charges and not by the conductor. The conductor simply provides an initial pathway for the charges to take up their position. It is similar to the case of a charged ring. if a test charge is placed at the center of a charged ring, it will experience zero electric force because the force due to any element of ring is canceled by the force due to element on the opposite side of the ring (because of symmetry) and if the force is zero, the electric field at the center of the ring would have to be zero.

However it can also be understandable by simply assuming a Gaussian surface inside the conducting shell. No charge is ENCLOSED by the surface so no electric field.

$\endgroup$
0
$\begingroup$

http://www.feynmanlectures.caltech.edu/II_05.html#Ch5-S10

This answers the question almost completely, unless you want a fully rigorous approach. The only intuition is that the inside surface cannot contain a purely positive/negative charge distribution. Therefore, there must be a region on the surface where the charge goes for + to -. Drawing a loop skimming this region, we can intuitively see why this solution works by the Kelvin-Stokes theorem.

$\endgroup$

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.