# Conceptual explanation of electrostatic shielding

If you put a closed conducting surface in any external electric field, the field inside the conducting surface will be zero.

A similar effect occurs if one has charged conducting surface without an external electric field. For that case you can argue on a conceptual level that the force on a test charge inside of the surface will be zero by using Coulomb's law (here a picture from Rogers: "Physics for the inquiering mind"):

You can sweep the whole spere with pairs of cones as indicated. Since the force acted by each particle on the test particle is proportional to the inverse square of the distance (Coulomb's law) and because the number of charges inside of the bottom of a cone is proportional to the distance squared, by superposition both contributions cancel out, so that the charges on area $A_1$ act on the test charge with the same but opposide force as that one from $A_2$.

Now I am looking for a similar (in style and level) argument of how to explain my first statement, i.e. a non charged closed surface in an external electric field. Let the surface be a sphere (or general) and the field inhomogenic (using a square box and a homogenous field this is easy to).

Edit As suggested in a comment, it the freely moving charges mus rearrange in such a way that the resulting field inside will be zero. I think it is conceptually clear that they rearrange in some way because the external field acts on them. However the point is, why do they reaarange in the "right way", i.e. such that the field inside is zero.

• The external electric field causes rearrangement of charges of the conducting surface such that the field inside the surface is zero. – Yashas Jan 26 '17 at 9:22
• @YashasSamaga: I can conclude that from observing that the field is zero inside. However how can I explain, that the reconfiguration will indeed in such a way that the field inside is zero? – Julia Jan 26 '17 at 12:01