# Spherical conducting shell

by Deoxygenation
Tags: conducting, shell, spherical
 P: 19 Spherical conducting shell It doesn't. The easiest way to think of it is by Gauss' Law (for electrostatics, obviously). Take the Gaussian surface to be a sphere outside the conducting sphere. If the charge in the middle is Q, and the charge on the conducting sphere is 0, then the flux ($$\Psi$$) through the Gaussian sphere is Q (total enclosed charge). Then, since $$\Psi = \oint \vec{D} \cdot d \vec{S} = Q \neq 0$$ Then, D (and E) can't be zero. Generally speaking, shields are grounded, though. If this one were grounded, then, while originally neutral, it would take on a charge of -Q once the Q charge is inserted in the sphere, the, by Gauss' Law, the external field is 0. I'll let you think about this one, but if you have a region with some field E, then put a floating conducting sphere in it, the field inside the sphere would be 0 and the field elsewhere would be unchanged, so shielding kind of works then.
 P: 550 To visualize it more easily, imagine a uniform electric field in which an uncharged metal shell is placed. Let's say the electric field points from left to right. This means that positive charges (or lack of negative charges, electrons) in the metal shell will start moving to the right and negative charges start moving to the left. After a short while, the left of the shell will be negatively charged (excess of electrons) while the right will be positively charged (lack of electrons). This will in turn generate it's own electric field, pointing from positive to negative, or right to left. The magnitude of this electric field (inside the spherical shell) is exactly the same as the magnitude of the field outside the shell, only in the exact opposite direction. The result is that, within the shell, there is NO net electric field. If you know Gauss' law you can easily check this: $$\oint \vec{E} \cdot d \vec{A} = \frac{Q_{\text{encl}}}{\epsilon_0}$$ If you take a Gaussian surface (spherical) inside the shell, you are not enclosing any charge and thus the electric field is 0.