Electric field inside & outside of a spherical shell

[mondo]

Hi,

I am reading Griffiths Introduction to electrodynamics. Currently I am solving problem 2.11 which asks to find an electric field inside and outside a spherical shell of radius R.
Inside:
$$\int{E \cdot da} = \frac{Q}{e_0} = |E|4\pi r^2 = \frac{Q}{e_0} = 0$$ The result is $$0$$ because we have no charge inside the Sphare
Outside:
$$\int{E \cdot da} = \frac{Q}{e_0} = |E|4\pi R^2 = \frac{\sigma4\pi R^2}{e_0} => E = \frac{\sigma R^2}{e_0r^2}$$
And here is my question, for the electric field outside of the sphere we see it depends on how far away we are from the sphere ($$R$$) and it looks like it will grow with square of the distance from the sphare! Does it make sense? Shouldn't it get weaker and weaker as we move away from the charge distribution that is on the sphere?

Thank you.

 

[Ibix]

Isn't $R$ the radius of the shell and $r$ the distance from the centre? That would make sense, giving the total charge $Q=4\pi R^2\sigma$ and hence $E=Q/4\pi\epsilon_0 r^2$.

 

[mondo]

I think you are right. Thank you.
So in light of that, the electric field diminishes in proportion to $$\frac{1}{r^2}$$ which makes sense.

There is one more thing that is not entirely clear to me. On one hand I agree that the electric field inside the sphere is 0 due to no charge inside. However the charge distributed on the surface of the sphere should generate an electric field inside the sphere as well - in the same way as it does outside of the sphere. What do I miss here?

 

[PeroK]

There is one more thing that is not entirely clear to me. On one hand I agree that the electric field inside the sphere is 0 due to no charge inside. However the charge distributed on the surface of the sphere should generate an electric field inside the sphere as well - in the same way as it does outside of the sphere. What do I miss here?

That field cancels itself out. This is one part of the Shell Theorem.

 

[Ibix]

What do I miss here?

It should be obvious that the field is zero at the centre of the sphere, because the contribution from any point on the sphere is cancelled by the contribution from the opposite point. (Remember $E$ is actually a vector, $\vec E$.)

It's less obvious off center, but roughly speaking as you move away from the centre there is less charge on one side of you than the other, but it's closer to you. The reduced average $r^2$ and the reduced $q$ cancel out. You can find plenty of formal proofs of the Shell Theorem online. Wikipedia's looks OK at a glance.

 

[Orodruin]

I think this is one of those cases where it is actually instructive to perform the integral of all contributions and see that they cancel out even if we know that they must simply on the basis of Gauss’ law.

Hint: The integral for finding the potential rather than the field is a bit easier. It should result in a constant expression inside the shell.

Edit: In other words, take a point $\vec x’$ and compute the integral
$$
V(\vec x’) = \int_{r=R} \frac{\sigma}{4\pi\varepsilon_0 |\vec x - \vec x’|} dS
$$

 

[Vanadium 50]

I think this is one of those cases where it is actually instructive to perform the integral

I agree. There is too much "I am not going to do the calculation, but think I would get a different result than in the book."

 

[Meir Achuz]

Use Gauss's law.