# Non-Zero Electric Field Inside A Conducting Shell

Statement:

"The electric field is zero everywhere inside the conductor, whether the conductor is solid or hollow."

Question:

A coaxial cable consists of a long, straight filament surrounded by a long, coaxial, cylindrical conducting shell. Assume charge $Q$ is on the filament, zero net charge is on the shell, and the electric field is $E_1\hat{i}$ at a particular point $P$ midway between the filament and the inner surface of the shell.

Next, you place the cable into a uniform external field $2E \hat{i}$. What is the $x$ component of the electric field at P then?

(a) 0

(b) between 0 and E1

(c) E1

(d) between 0 and 2E1

(e) 2E1

Answer (c). The outer wall of the conducting shell will become polarized to cancel out the external field. The interior field is the same as before.

My Problem:
Wasn't it suppose to be 0 inside a conductor in whatever condition while it is in equilibrium? How come is it a non-zero value? I would expect the conducting shell to cancel out the electric field within it. That's what I've seen in the previous questions I've solved. Please explain.

• The first statement is 100% false, even though it can be found in many poorly written textbooks. Commented Apr 19, 2016 at 14:19
• Then tell me the correct one instead. Commented Apr 19, 2016 at 14:23
• "The electromagnetic field inside a hollow conductor can be whatever (Maxwell's equations allow)." Commented Apr 19, 2016 at 14:24
• The statement should be that the electric field is zero in the 'flesh' of a conductor, whether it is solid or hollow, in the case of electrostatics. The field inside any cavities inside a conductor may be nonzero.
– Jeff
Commented Apr 19, 2016 at 14:44
• @Jeff: Why did you have to go & answer right after the edit window closed? ;-) Commented Jul 17, 2017 at 18:59

With reference to your question, if in case 1 , the field was $E_{1}\vec{i}$, then in case 2 it would remain the same because electric field lines of external field and the polarised charges of the outer surface would not reach the inside of the conductor(as can be seen from diagram depicting field lines). Thus the RESULTANT( not the contribution due to individual), of external and outer surface charge would cancel each other.
Now if we take a cylindrical gaussian surface just enough to enclose the inner surface charges, since we know that field inside MATERIAL of conductor is zero, thus application of gauss law gives net charge enclosed as zero, which further yields that inner surface must have uniform distribution of $-q$ charge( uniform only because the filament is AT CENTER, if it would have been off center, distribution would be non uniform), thus by invoking symmetry arguements(or gauss law, if you want) we get that field due inner surface charge is also zero. Therefore filed will be only due to filament, thus case 2 is equivalent to case 1.
All the points on the inner surface of the hollow conductor have the same potential. Therefore, if the charge density $\rho$ inside is zero everywhere, the electrostatic field will be zero. In your case the filament carries a charge $Q$, so the charge density is not zero everywhere.