Understanding the Electric Field of Two Spheres: A Scientific Approach

[Feeziks]

Homework Statement

A sphere of radius r is uniformly charged with volume charge density 𝜌. A
hollow conductive sphere with internal radius r and external radius R is tightly
wrapped around the first sphere, and it has a total charge Q. Find the electric field
in any point as a function of the distance from the centre.
[You have to use Gauss law. For the conductor, be careful: you first need to find
out how much charge is on the internal surface and how much is on the external
one]

Relevant Equations

Gauss Law: Φ = Q / ε = E * dA

I am not quite sure how to present my answer in the form of a function with relation to the distance from the centre.

What I got so far is the E1 and E2, for the internal and external sphere respectively.

For internal sphere, the charge is volume * 𝜌, so it is
$$ \frac{4\pi r^{3}}{3} * 𝜌$$
Now let this be q. I know that E *Area = q / ε, and area of a sphere is $$ 4 \pi r^2 $$
so E1 = q /(ε*Area), which gives $$E1 = \frac{r𝜌}{3ε} $$

I do the same for the external sphere, and I get $$E2 = \frac{Q}{ε4\pi R^2} $$

So now I know E1 and E2, great. If the distance from the centre is less than r, E will be 0. The moment the distance is greater than r but less than R, I just refer to E1. But the moment the distance is greater than both r and R, I need to account for E1 and E2. Do I just sum up both E1 and E2? In this case, shouldn't there be 2 different functions in place?

 

[BvU]

Hello @Feeziks,

!​

So now I know E1

Agreed. But using the symbol $r$ for radial distance is misleading: $r$ is the radius of the inner sphere.

I do the same for the external sphere, and I get

That is not correct.

How big is E in the range $(r, R)$ ?

For the conductor, be careful: you first need to find
out how much charge is on the internal surface and how much is on the external
one

Did you do anything with this valuable information ?

 

[Feeziks]

Did you do anything with this valuable information ?

Ah, I thought that by dividing the E into E1 and E2 I had used that valuable information. I guess not. Let me try again.

Also, for

But using the symbol $r$ for radial distance is misleading: $r$ is the radius of the inner sphere.

I thought that both the volume and the area utilises the same r? Since the internal sphere is fixed, so the volume and area should use the radius, r of the internal sphere. Unless the "area" is actually referring to that of the "sphere" of "r*" which is the function of the distance from the centre? I think as long as r* is less than r, the electric field will be equivalent 0?

How big is E in the range $(r,R)$?

In the range of (r,R), I believe that only E1 comes into play, since it is still inside of the external sphere. However, once the distance is greater than R, we need to account for E2, so the total electric field once r* exceeds R will be a function of E1 and E2, am I right?

 

[haruspex]

I know E1 and E2

That is not correct.
...
using the symbol r for radial distance is misleading: r is the radius of the inner sphere.

I think @Feeziks is defining E2 to be the field due to the conductor only, just outside it. Similarly E2. r is not being used as a variable.

If the distance from the centre is less than r, E will be 0.

No, this is not a conductor. At radius s from the centre, consider separately the field due to the charge at radius less than s and that due to the charge at radius greater than s.

The moment the distance is greater than r but less than R, I just refer to E1.

You are overlooking this hint:
you first need to find out how much charge is on the internal surface

 

[BvU]

I think as long as r* is less than r, the electric field will be equivalent 0?

Using $r^*$ for the radial distance is fine. How much charge is enclosed by a gaussian spherical surface with radius $r^*<r\$ ?

Let me try again.

Good plan.

In the range of (r,R), I believe that only E1 comes into play

In the range $(r,R)$ you are inside a conductor. What does that mean for $\vec E$ ?

 

[Feeziks]

" No, this is not a conductor. At radius s from the centre, consider separately the field due to the charge at radius less than s and that due to the charge at radius greater than s. "
"You are overlooking this hint:
you first need to find out how much charge is on the internal surface "
"Using r* for the radial distance is fine. How much charge is enclosed by a gaussian spherical surface with radius r* < r?"
" In the range (r,R) you are inside a conductor. What does that mean for E? "

Let me take into account the comments and what I have gathered online (including https://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Fields-and-Conductors --> specifically the part on "Electric Fields Inside of Charged Conductors") to refine my working.

The charge in the internal sphere is
$$ \frac{4\pi r^{3}}{3} * 𝜌$$
Which is q. This is uniformly charged as stated.
The electric field due to this charge q is $$Eq = \frac{q}{ε4\pi r*^2} $$
where r* varies according to the distance from the centre.
So Eq = q /(ε*Area), which gives $$Eq = \frac{r^3𝜌}{3εr*^2} $$

So the above is for when r* < r. I hope this is right now...

And moving on, when in the range (r,R), it is in a conductor so E should be 0! Am I correct?

So when 0 < r* < r, we look at Eq. But once r < r* < R, E is 0. Then once r* > R, we look at E2 as per above. I am guessing that the internal sphere does not matter anymore because this is a form of "shielding"?

Edit: Please do correct me if I made any conceptual errors!

 

[BvU]

And how did you now take the urgent hint into account?

Which is q. This is uniformly charged as stated.
The electric field due to this charge q is $$Eq = \frac{q}{ε4\pi {r^*}^2} $$
where r* varies according to the distance from the centre.

Should be (check the dimensions) $$E = \frac{q}{ε4\pi {r^*}^2} $$ ($E$, not $Eq$). Only for $\ r^* > r \$ and only if nothing else is sitting there.
So not for $\ 0 < r^* < r$

correct me if I made any conceptual errors!

Introducing $q$ only confuses you. Stick to using $\rho$.
Consider a spherical gauss surface with radius $\ 0 < r^* < r$

In the range $(r,R)$ you are inside a conductor. What does that mean for $\vec E$ ?

And moving on, when in the range (r,R), it is in a conductor so E should be 0! Am I correct?

You got the message. Now think about

For the conductor, be careful: you first need to find out how much charge is on the internal surface and how much is on the external one]

Consider a spherical gauss surface with radius $\ r < r^* < R$

Then once $r^* > R$ , we look at E2 as per above. I am guessing that the internal sphere does not matter anymore because this is a form of "shielding"?

Edit: Please do correct me if I made any conceptual errors!

So if $Q=0$ your reasoning is that there is no field outside the spheres ? In other words poor old Gauss is dead wrong ? No way ! Don't guess, but think again !

The conceptual error is: the shielding provided by conductors shields the inside from outside fields. Not the other way around. Gauss is right!

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[Feeziks]

Ok, let me try that again. So when r* > r *and* there is no other items around it, the internal sphere's E field is
$$E = \frac{q}{ε4\pi r*^2} $$
This is the case where the external conducting sphere does not exist. However, the external hollow sphere exists, so E is not as above. I need to use the hint given.

And now let's assume now that r* < r.
So the total charge within = $$ q\frac{r*^3}{r^3} $$
It is to the power of 3, because the volume of sphere has radius to the power of 3. q was the original charge. This means that the charge inside should be less than q. This will be a linear function, so the E field increases linearly when r* < r.

And the moment r* > r, the E field will be inversely related to r squared. So the further away I go, the weaker the E field.

All the above is assuming there is no external hollow conductor...yet.

For the hint " how much charge is on the internal surface ", I believe the internal surface refers to r < r* < R as mentioned, where I should think about a spherical Gaussian surface.
So for the charge on the internal surface where r < r* < R, I would first take E as above, as in $$E = \frac{q}{ε4\pi r*^2} $$, because it would be influenced by the internal sphere only. The external sphere is a conductor, so E = 0. Then I know from Gauss Law that Φ = Q / ε = E * Area, so the charge is E * Area * ε.

And once r* > R, I will take the E field to be the sum of internal sphere + external sphere?

Am I going in the right track?

 

[BvU]

This means that the charge inside should be less than q. This will be a linear function

The charge inside will be a third order function !

Introducing $q$ only confuses you. Stick to using $ρ$.

At least for the inner sphere.

So for the charge on the internal surface where r < r* < R, I would first take E as above, as in $$E = \frac{q}{4\pi\varepsilon\, {r^*}^2} $$, because it would be influenced by the internal sphere only.

No. The field inside a conductor is 0. If not, charge moves until it is -- that's why it is called a conductor.

'because it would be influenced by the internal sphere only' is not correct. It would be influenced by anything up to $r^*$. In particular: any charge on the inner surface of the conducting outer shell.

Consider a spherical gauss surface with radius $\ r < r^*+\delta < R$

with $\delta << R-r$. You know the field, so you know the charge inside. Now you can bring in $q$ if you want, and you are able to answer

" how much charge is on the internal surface "​

$\$

 

[BvU]

Lost interest ?

 

[haruspex]

I believe the internal surface refers to r < r* < R

No, it refers to the inner surface of the conducting shell, i.e. radius r.
You know what the net field is just inside that, and you know what the net field is just outside that, so you can deduce that layer of charge.

The charge inside will be a third order function !

Judging from the surrounding text, @Feeziks meant the field inside the inner sphere will be linear with r*, which is true.

 

[BvU]

It is to the power of 3, because the volume of sphere has radius to the power of 3. q was the original charge. This means that the charge inside should be less than q. This will be a linear function, so the E field increases linearly when r* < r.

It worried me ...

OP was Last seen Wednesday, 5:28 PM, so he/she hasn't seen it yet

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