In a hollow conductor and in a spherical shell the charges reside on its surface and there is no electric field inside it.Please tell me the reason behind it.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ possible duplicate of physics.stackexchange.com/q/55119 $\endgroup$– user81619Oct 4, 2015 at 11:07
-
$\begingroup$ More on charges in a conductor. $\endgroup$– Qmechanic ♦Oct 4, 2015 at 11:28
-
5$\begingroup$ There are many duplicates. e.g. physics.stackexchange.com/q/109556 physics.stackexchange.com/q/22773 $\endgroup$– ProfRobOct 4, 2015 at 11:29
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
In all conductors, charges reside on the surface.
The reason for this is that conductors have free electrons, that is, the electrons are loosely attached to the nucleus of the atoms in the conductors.
Refer to the pic below (drawn in MS Paint)
When placed in external electric field, the electrons migrate to one side of the conductor and an electric field is created inside the conductor itself due to separation of positive and negative charges, as is evident from the pic. In case of conductors, this electric field is equal to that of the external electric field and hence neutralises the external field.
So the net field inside the conductor is zero.
answered Oct 4, 2015 at 11:35
-
$\begingroup$ Consider the case in which the external electric field is extremely strong. After all the free electrons migrate to one side of the conductor, if the electric field created inside the conductor itself due to separation of positive and negative charges is not so strong and cannot be equal to the external field, what happens? $\endgroup$– 佐武五郎Apr 3 at 9:28
-
$\begingroup$ I asked my question in the following link: physics.stackexchange.com/q/808892/374591 $\endgroup$– 佐武五郎Apr 3 at 23:41