Electric field lines outside a neutral conducting spherical shell (2024)

I'm talking about a situation like this. Ive been told that charge distribution on the inner side of the conducting material is non uniform and equal to -q. This makes perfect sense

But ive also been told that the charge distribution on the outer part of the conducting material is ##uniform##.
Can someone give me a valid argument for this? I saw explanation involving some difficult theorems like "uniqueness theorem". I do not understand them fully because of the advanced math involved and its not taught in my coursework (i am in 12th grade)

Further ive also come across a question where they ask us to find the field just outside the sphere. The answer to that is ##\frac{q} {4π\epsilon_{0}(2R)²}##
which I dont get why..

Note that

1. The electric filed is zero inside the conducting material in the region ##R\leq r \leq 2R.##
2. By Gauss's law, this means that the total charge induced on the inner surface at ##r=R## must be ##-q##.
3. If you start with a neutral conductor, the total charge on the outer surface must be ##+q##.
4. The charge on the outer surface can move around in response to changing electric fields.

Now here is the key argument that answers your question: Suppose you move charge ##+q## in the cavity to a new position. What will happen? Total charge ##-q## on the inner surface will redistribute itself to cancel the field inside the conductor according to (1) above. Since the field is zero before the charge in the cavity is moved and also zero after this charge is moved, the charges on the outer surface can see no change and, in fact, are not "aware" that there is charge inside the cavity. All they "know" is that they are on a spherically symmetric surface, they will distribute themselves accordingly and stay there until charges external to the conductor are brought nearby. Therefore, the electric field outside will be that of a uniformly charged sphere of radius ##2R.##

By the way,

tellmesomething said:

Further ive also come across a question where they ask us to find the field just outside the sphere. The answer to that is ##\frac{q} {4π\epsilon_{0}2R}##
which I dont get why..

I too don't get why the expression you quoted is the field just outside the sphere. It is the electric potential on the surface of the sphere.

kuruman said:

Now here is the key argument that answers your question: Suppose you move charge ##+q## in the cavity to a new position. What will happen? Total charge ##-q## on the inner surface will redistribute itself to cancel the field inside the conductor according to (1) above. Since the field is zero before the charge in the cavity is moved and also zero after this charge is moved, the charges on the outer surface can see no change and, in fact, are not "aware" that there is charge inside the cavity. All they "know" is that they are on a spherically symmetric surface, they will distribute themselves accordingly and stay there until charges external to the conductor are brought nearby. Therefore, the electric field outside will be that of a uniformly charged sphere of radius ##2R.##

This makes so much sense now. Thanks a ton.

kuruman said:

I too don't get why the expression you quoted is the field just outside the sphere. It is the electric potential on the surface of the sphere.

Sorry I missed the " ² ". But the argument you mentioned clears this one as well. Thankyou again.

tellmesomething said:

Further I've also come across a question where they ask us to find the field just outside the sphere. The answer to that is ##\frac{q} {4π\epsilon_{0}(2R)²}##
which I don't get why..

LaTeX tip:

Please do not use special UNICODE characters such as exponents, Greek characters, etc.

Rather than using ² for an exponent, use LaTeX formatting. Use " ^2 " or " ^{2} ".

##\frac{q} {4π\epsilon_{0}(2R)^2}\ ## is easier to read than ##\ \frac{q} {4π\epsilon_{0}(2R)²}\ ##.

kuruman said:

Now here is the key argument that answers your question: Suppose you move charge +q in the cavity to a new position. What will happen? Total charge −q on the inner surface will redistribute itself to cancel the field inside the conductor according to (1) above. Since the field is zero before the charge in the cavity is moved and also zero after this charge is moved, the charges on the outer surface can see no change and, in fact, are not "aware" that there is charge inside the cavity. All they "know" is that they are on a spherically symmetric surface, they will distribute themselves accordingly and stay there until charges external to the conductor are brought nearby. Therefore, the electric field outside will be that of a uniformly charged sphere

I don’t think that quite does it.
You have shown that if +q at point P has a solution in which the internal and external distributions are ##D_i, D_o## then with +q at point P' there is a solution in which ##D_o'=D_o##. The challenge is to show it is the only solution.
To put it another way, total charge −q on the inner surface could redistribute itself to cancel the field inside the conductor, but maybe there is a way in which it does not do that by itself and instead it is achieved by the combination of how the internal and external charges redistribute.

haruspex said:

The challenge is to show it is the only solution.

Yes, but

tellmesomething said:

I saw explanation involving some difficult theorems like "uniqueness theorem". I do not understand them fully because of the advanced math involved and its not taught in my coursework (i am in 12th grade)

I tried to make it simple and commensurate with OP's level.

kuruman said:

I tried to make it simple and commensurate with OP's level.

Ok, but we should not leave @tellmesomething with the impression that no equivalent to the uniqueness theorem is needed for an actual proof.