Re: Surface charge distribution

[Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

"Fakhruddin, Hasanbhai" wrote:

> Both the replies agree that the charge distribution on the outer surf=
> ace
> will be uniform.  However, I still have some questions:  =20
>
> 1. If the point charge inside is +q will the total charge on the oute=
> r
> surface be +q (and the charge on the cavity wall be -q)?
>
> 2.  Assume that the point charge is close to the cavity surface.  Now=
>  if
> we increase the radius of the sphere (keeping the point charge at the
> same distance from the cavity surface) the charge density on the oute=
> r
> surface will decrease (right?) When the radius becomes infinity the p=
> art
> of conducting sphere close to the point charge becomes a conducting
> plate.  The E-field on the other side of the plate will become ZERO
> because the charge density on the plate will be zero!   Is that corre=
> ct?

Yes! The field outside the conducting spherical shell is the same as if
one place the original inner charge at the center of the sphere. As the
radius of the sphere increases, the outside charge is distributed over a
larger area (surface charge density decreases) and the measure fields have
to be small just because you are so far from the center. In the limit of
infinite radius, the surface charge density becomes zero and the field on
the "outside" is that of a charge infinitely far away at the "center",
i.e., it's zero.

Bob at PC