Question

How does Gauss Law work inside a conducting sphere?

A point charge +Q is inside an uncharged
conducting spherical shell that in turn is near
several isolated point charges, as shown above.
The electric field at point P inside the shell
depends on the magnitude of
(A) Q only
(B) the charge distribution on the sphere only
(C) Q and the charge distribution on the sphere
(D) all of the point charges
(E) all of the point charges and the charge
distribution on the sphere

The correct answer is (A) and the explanation given is due to Gauss Law.

The derivation of the Gauss Law is quite complicated for me to understand. But isn't the electric flux in Gauss' only include the flux as the result of the enclosed surface? How could the electric field at point P not depend on `q_1`, `q_2` and `q_3`?

I mean choosing the enclosed surface to be inside the conducting sphere and then conclude that therefore the electric field at `P` only depends on `Q` seems a bit shady to me. Couldn't we choose a big enough closed surface to include the three external charges as well? What does Gauss Law say exactly?

Answer 1

See below

Explanation:

Agreed.

Consider the simplest example of 1 charge, +Q. If we draw an imaginary sphere concentric with +Q , Gauss' Law tells us that:

• "The net electric flux out of a closed surface - our sphere - is equal to the charge enclosed, ie +Q, divided by the permittivity."

Or:

• `Phi_(n\et) = int int_A bb E cdot d bb A = Q_(enc)/epsilon_o`

The great trick with Gauss' law is to exploit some given symmetry.
We can argue that symmetry demands that the `bb E` field at any point on our spherical surface is the same and points outward orthogonally to our sphere (which has radius `r`).

Ie:

• `int int_A bb E cdot d bb A =color(red)(E ( 4 pi r^2)) = Q/epsilon implies E = Q/(4 epsilon pi r^2)`

Et voilà, Coulomb's Law!

Now add a further charge +q, placed some short distance from +Q, but not within the imaginary sphere.

The prediction made earlier by Gauss' Law about net flux is still true. Flux due to +q is entering but then leaving our imaginary sphere. It all nets out so the net flux is all due to +Q.

But, crucially, there is no neat symmetry. We cannot make any simplifying assumption about the `bb E` field about +Q. We need to superpose the individual `bb E` vector fields: `bb E_("net") = bb E_("+Q") + bb E_("+q")`

For example: `Q " << " q implies bb E_("net") approx bb E_("+q")`

Finally, now include the spherical conducting shell, enclosing +Q.
Outside that shell there is +q and any number of other charges.

Both +Q and the external charges will cause a charge re-distribution within the conducting shell, the effect of which is to minimise the electrical potential energy of the entire system .

The nature of that distribution will be driven by the locations of +Q and the outside charges. Crucially it will not be symmetric except in certain circumstances not described in the Question you have shared. It is only with perfect geometric and charge symmetry that we can use Gauss to conclude that the `bb E` field is zero in a conducting shell .

Again Gauss' Law will be true: "The net electric flux out of our sphere will equal +Q, divided by the permittivity."

But again, the field inside will be a superposition of any number of other `bb E` fields.

So the answer is E.