# Charge distribution in concentric shells

## Homework Statement

A solid conducting sphere of radius 2.00 cm has a charge 16.00 µC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a total charge of -3.00 µC. (Take radially outward as the positive direction.)
Find the electric field at r = 4.50 cm from the center of this charge configuration.

## The Attempt at a Solution

I think there's something wrong with my physical picture. So before I get into any calculations, this is what I'm thinking.

Since they are conductors the charge can move around. The inner sphere is positive so it will attract the negative charges from the outside of the outer sphere to the inside, leaving the outside of the outer sphere positive.
Therefore, the net charge within the enclosed Gaussian should be Qin=+13μC.

However, that doesn't seem to yield the correct answer.

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Are you sure the question states that the outer shell is conducting ? If yes , then the answer is pretty simple . What is the electric field inside a conductor ?

Yes I'm sure.

It's 0 inside a conductor. But the Gaussian is also surrounding the smaller conducting sphere, that should account for some field?

BvU
Homework Helper
2019 Award
Yes, all the way to the inner radius of the outer shell, where you have correctly drawn a bunch of +++ ! Just a skin depth into the outer shell the field really is 0 (otherwise more charges would move until it is).

Wait, I didn't understand that. Can you explain again?

Yes I'm sure.

It's 0 inside a conductor. But the Gaussian is also surrounding the smaller conducting sphere, that should account for some field?
The diagram you have made has incorrect amount of negative charge -3.00 µC on the inner surface of outer shell .

If you know electric field is zero within the conductor then you are done with the problem .

Even if you wish to apply Gauss Law to this problem ,you need to use the fact that electric field inside a conductor is 0 .

Take a gaussian sphere of radius 'r' , 4<r<5 .Since E = 0 , ∫E.ds = 0 which as per Gauss Law implies charge enclosed in the gaussian sphere is 0 .This in turn means that the charge enclosed on the inner surface of the outer shell is -16µC .The charge on the outer surface of outer shell will be 13 µC ,such that the net charge on the outer shell is -3.00 µC .The electric field lines from the outer shell are radially outwards .

The electric field lines originate from the outer surface of inner sphere and terminate at the inner surface of outer shell such that the electric field at r , 4<r<5 is zero .

1 person
BvU
Homework Helper
2019 Award
... have correctly drawn a bunch of +++ ! ..
Sorry, that was a bunch of --- (I didn't look at the number -3, but Tanya handled that, and the remainder excellently).