The Electric Field Due to a Thin Spherical Shell, Gauss's Law

[swraman]

1. A thin spherical shell of radius a has a total charge Q distributed
uniformly over its surface. Find the
electric field at points outside the shell and inside the shell



2. \intE dA = Q/\epsilon



3. THis is an example in my textbook, and it claims that:

The calculation for the field outside the shell is identical
to that for a solid sphere (which yields E = KQ/r^2). If we
construct a spherical gaussian surface of radius r >a concentric
with the shell, the charge inside this surface
is Q. Therefore, the field at a point outside the shell is equivalent
to that due to a point charge Q located at the center.

I don't see how it is equivalent to a point charge at the center, or how it is the same as the calculation for a sphere.

The electric field of a sphere with total charge > 0 is directed directly outward from the sphere and thus will be perpendicular to a gaussian surface (as long as the centers are the same).
But in this case of a ring, the electric field will not be perpendicular to the gaussian surface all the time, will it?

Thanks

 

[borgwal]

"But in this case of a ring, the electric field will not be perpendicular to the gaussian surface all the time, will it?"

A spherical shell is not a ring.

 

[swraman]

"But in this case of a ring, the electric field will not be perpendicular to the gaussian surface all the time, will it?"

A spherical shell is not a ring.

yea...jsut noticed that...now I feel stupid.

thanks