Electric field lines outside a neutral conducting spherical shell

[tellmesomething]

Homework Statement

I could not add the full thing in the title due to word limit, apologies. Electric field lines due to a neutral conducting spherical shell with a point charge placed offcentrically inside

Relevant Equations

None

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..

 

[kuruman]

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,

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.

 

[tellmesomething]

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.

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.

 

[SammyS]

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}\$.

 

[haruspex]

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.

 

[kuruman]

The challenge is to show it is the only solution.

Yes, but

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.

 

[haruspex]

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.

 

[tellmesomething]

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.

I just came across a variation of this question so i have a little doubt if you dont mind clearing:
Suppose we have a charged conducting spheres(radius R) with the same scenario as above I.e a charge placed off centrically,
Suppose we keep another point charge outside the sphere
Take the point charge inside to be q
Charge on spherical shell to be q
Charge outside to be 2q

If there was no charge outsider charge on inner surface of surface of sphere would be
-q(non uniform)
and q+q on the outer surface of sphere (uniformly distributed)

Now due to this 2q charge how would the situation change?
I dont exactly get It, my teacher says that due to the q point charge the outer surface would still have a q uniform distribution

But due to the 2q placed outside the original charge of the spherical charge (q) would be non uniformly distributed.
Hence the outer surface would have q(uniform)+q(non uniform)=2q(non uniform)
This seems odd to me . I expected two different things
i)Charge on the outer surface to be non uniform for both of the q's
ii) charge on the outer surface to be -2q as well like its for the charge on the inner surface due to +q point charge..

 

[kuruman]

Before I answer, I need to understand what you are saying.

• You have charge $+q$ inside the cavity at $r<R.$
• You have net charge $+q$ on the spherical shell.
• You have charge $+q$ in the space beyond the spherical shell at $r>2R.$
• You want to know how the charges are distributed on the shell.

Is all that correct?

 

[tellmesomething]

• .
• You have charge $+q$ in the space beyond the spherical shell at $r>2R.$

Is all that correct?

All correct except this, the charge at $r$ where $r>2R$ is $+2q$

 

[kuruman]

Since there is charge $+q$ inside at $r<R$, there will be total charge $-q$ at $r=R$ on the inner surface of the shell. This means means that there will be total charge $+2q$ on the outer surface at $r=2R$ to keep the net charge on the shell at $+q.$ Because there is additional charge outside the shell at $r>2R$ (it doesn't matter how much and of what sign), the charge distribution at $r=2R$ will not be uniform but it will still add up to a total charge $+2q.$

 

[tellmesomething]

Since there is charge $+q$ inside at $r<R$, there will be total charge $-q$ at $r=R$ on the inner surface of the shell. This means means that there will be total charge $+2q$ on the outer surface at $r=2R$ to keep the net charge on the shell at $+q.$ Because there is additional charge outside the shell at $r>2R$ (it doesn't matter how much and of what sign), the charge distribution at $r=2R$ will not be uniform but it will still add up to a total charge $+2q.$

And the magnitude of outer surface charge is not affected by the outer point charge? Only the distribution on the outer surface is affected ?

 

[kuruman]

And the magnitude of outer surface charge is not affected by the outer point charge? Only the distribution on the outer surface is affected ?

Yes.