Still more problems with Gauss Law

1. Oct 6, 2004

stunner5000pt

If a spherical cavity of radius 3.66cm in a piece of metal (kind of cube, but not a perfect cube) has a charge of +Q at it's centre and there is a point P1 located half way between the spherical cavity and it's surface and a point P2 located in the metal piece , use gauss Law to find the electric field at
a) Point P1
b) point P2

a) This is fine for me

Let e0 = permittivity of free space epsilon 0

e0 SurfaceIntegral (E dA) = q
e0 E (4 pi r^2 ) = q
E = kq /r^2
which is coulomb's law derived! and just plug and away i go!

b) But for point P2

err can i actually use a spherical surface here??

as in do the same thing as before but use a bigger value for r i.e. radius of sphere + distance of sphere to point P2

appropriate? or not?

2. Oct 6, 2004

Gokul43201

Staff Emeritus
Umm...the answer to that would be : NOT YET. Mostly because you don't yet know what charges are enclosed by this Gaussian surface.

Read what your text has to say about induced charges and electric fields inside a piece of metal.

3. Oct 7, 2004

Claude Bile

You can still use a spherical surface though.

Claude.

4. Oct 7, 2004

stunner5000pt

i'm not quite sure how i would use it , so the shere would have a radius of 3.66cm + A ??

e0 E 4 pi (3.66+A)^2 = 126 x 10^-9

something liek this??

5. Oct 10, 2004

Claude Bile

You need to take into account the induced charges on the inner surface of the metal.

Claude.