Calculating electric fields

[ibaraku]

Homework Statement

Two concentric plastic spherical shells carry uniformly distributed charges, Q on the inner shell and -Q on the outer shell. Find the electric field (a)Inside the smaller shell, (b)between the shells, and (c) outside the larger shell

Homework Equations

integration(E * n)dA = (4pi) (k) (Qenclosed)

The Attempt at a Solution

In order to get hteh electric field between the shells, we can say that

[abs(E) * abs(n) cos delta](4pi) r^2 = (4pi) (k) (Qenclosed)

and working out the algebra it comes out to be

E = kQ/pi r^2

but what about for the charge outside the larger shell and the charge inside the smaller charge?
Is it safe to say that the we need to divide by 2 for the smaller shell and multiply by 2 for the large shell?
Thanks

 

[Defennder]

[abs(E) * abs(n) cos delta](4pi) r^2 = (4pi) (k) (Qenclosed)

and working out the algebra it comes out to be

E = kQ/pi r^2

Yes that's right.

but what about for the charge outside the larger shell and the charge inside the smaller charge?
Is it safe to say that the we need to divide by 2 for the smaller shell and multiply by 2 for the large shell?
Thanks

You do this the same way as you did the above, except now the Gaussian surface for (a) cuts through the smaller sphere, and in (c) the Gaussian surface encloses both spheres. There isn't any way you can use the result obtained above to compute for the other two cases. The method used is still valid, though because of symmetry.

 

[baba89]

for your question. For the charge outside the larger shell, the electric field can be found using the same equation, but with the enclosed charge being -Q instead of Q. So the electric field would be E = -kQ/pi r^2. As for the charge inside the smaller shell, we can use the same equation, but with the enclosed charge being 0, since there is no charge inside the smaller shell. So the electric field inside the smaller shell would be E = 0. It is not necessary to divide by 2 or multiply by 2 for these cases.