# Can charges move without a field? How charges redistribute without it?

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• Aurelius120
Aurelius120
TL;DR Summary
The problem arose when I tried to find the capacitance of two concentric shells when the inner one is earthed. For this it was necessary that the charge on the inner sphere be found. After which they can be treated as parallel combination of capacitors or something similar. However in the absence of an electric field how does charge induce on the inner sphere when a charge is stored on the outer?
So from Gauss theorem, electric field at any point inside a uniformly charged sphereical shell is zero. Thus there is no electrostatic force on the inner sphere.

From what I have learnt, a field is necessary to move charges. But in this case the inner sphere acquires a charge q without any force acting on it.

I fully understand the electrostatic potential explanation. That the Earthed sphere must maintain a zero potential therefore some charge flows into it from earth to make net potential zero, according to equation:
$$\frac{Kq}{r_1}+\frac{KQ}{r_2}=0\implies q=\frac{-Qr_1}{r_2}$$

In other cases of electrostatics, movement of charges is explained by field
Why does that fail here? Is field not necessary for motion of charges?

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Even in current electricity the motion of charges is explained by field developed due to emf of cell.$$I=nAev_d=nAe\left(\frac{eE}{m_e}\tau\right)$$
In this question as well, the explanation for why capacitor opposes current flow in charging is that the field for finite area charged sheet is non-uniform and therefore the field outside the capacitor is not zero so current flow is opposed. Even this can be explained by field

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Aurelius120 said:
TL;DR Summary: So I was asked to find the capacitance of two concentric shells when the inner one is earthed. For this it was necessary that the charge on the inner sphere be found. After which they can be treated as parallel combination of capacitors or something similar. However how does charge induce on the sphere?

So from Gauss theorem, electric field at any point inside a uniformly charged sphereical shell is zero. Thus there is no electrostatic force on the inner sphere.

View attachment 341854
From what I have learnt, a field is necessary to move charges. But in this case the inner sphere acquires a charge q without any force acting on it.

I fully understand the electrostatic potential explanation. That the Earthed sphere must maintain a zero potential therefore some charge flows into it from earth to make net potential zero, according to equation:
$$\frac{Kq}{r_1}+\frac{KQ}{r_2}=0\implies q=\frac{-Qr_1}{r_2}$$

In other cases of electrostatics, movement of charges is explained by field
Why does that fail here? Is field not necessary for motion of charges?

Even in current electricity the motion of charges is explained by field developed due to emf of cell.##I=nAev_d=nAe\left(\frac{eE}{m_e}\tau\right)##
The capacitance that you were asked to find is a geometric property. If you put charge on the spheres, the capacitance is not going to change.

That gives us the capacitance. A fixed value of capacitance means a charge corresponding to that must be on the inner sphere( ##\neq 0##) which got there without an electric field

Is it possible to ground the inner sphere inside a spherical outer conductor?

I am not sure what this question is expecting you to assume. To ground an inner conductor would seem to me to disrupt the symmetry. So I am not sure how much of it you are supposed to assume.

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PeterDonis
Dale said:
Is it possible to ground the inner sphere inside a spherical outer conductor?
Yes. You are standing on it.
That is what the Earth is, inside the ionosphere.

Dale
Aurelius120 said:
from Gauss theorem, electric field at any point inside a uniformly charged sphereical shell is zero.
If the inner sphere is grounded, the outer sphere cannot be a uniformly charged spherical shell and Gauss's theorem cannot apply.

In the video below the concentric shells are treated as a parallel combination of two spherical capacitors formed by three concentric shells. The third, outermost shell is at infinity, it is at zero potential, and forms a capacitor with the middle shell of radius ##b## which is at some potnetial ##V##. It has capacitance $$C_2=\frac{4\pi\epsilon_0}{\frac{1}{b}-\frac{1}{\infty}}={4\pi\epsilon_0~b}.$$ The middle shell of radius ##b## and inner shell of radius ##a## also form a capacitor ##C_1##. It is in parallel with ##C_2## because the inner shell is at the same potential as the shell at infinity and the middle shell is common to the two. Its capacitance is $$C_1=\frac{4\pi\epsilon_0}{\frac{1}{a}-\frac{1}{b}}=\frac{4\pi\epsilon_0~a~b}{b-a}.$$The equivalent capacitance is ##C_{eq}=C_1+C_2.##

The language in the video is incomprehensible to me but not the math. I showed the gist of the argument above for those who don't have the patience to sit through this 4 min video.

Aurelius120
PeterDonis said:
If the inner sphere is grounded, the outer sphere cannot be a uniformly charged spherical shell and Gauss's theorem cannot apply.
Why though? Why should grouunding the inner sphere cause any problem to the outer sphere? It's still a metal and therefore charge will dustribute to get to maximum distance from each other right?

Aurelius120 said:
Why though?
The conductor that grounds the inner sphere, must pass through a hole in the outer sphere.

PeterDonis
Baluncore said:
Yes. You are standing on it.
That is what the Earth is, inside the ionosphere.

Baluncore said:
The conductor that grounds the inner sphere, must pass through a hole in the outer sphere.
So a metallic Earth without a metallic Ionosphere is not possible.
Also won't the hole affect the potential?

This is how I was shown to find the potential and then capacitance:
$$\frac{KQ}{b}+\frac{Kq}{a}=0\implies q=\frac{-Qa}{b}$$
Then potential difference$$=\frac{KQ}{b}-\frac{KQa}{b^2}$$
Then assime parallel combination of capacitors OR
The charge stored on capacitor is Q for voltage above.
The capacitance is $$\frac{4\pi\epsilon_○b^2}{b-a}$$

kuruman said:
In the video below the concentric shells are treated as a parallel combination of two spherical capacitors formed by three concentric shells. The third, outermost shell is at infinity, it is at zero potential, and forms a capacitor with the middle shell of radius ##b## which is at some potnetial ##V##. It has capacitance $$C_2=\frac{4\pi\epsilon_0}{\frac{1}{b}-\frac{1}{\infty}}={4\pi\epsilon_0~b}.$$ The middle shell of radius ##b## and inner shell of radius ##a## also form a capacitor ##C_1##. It is in parallel with ##C_2## because the inner shell is at the same potential as the shell at infinity and the middle shell is common to the two. Its capacitance is $$C_1=\frac{4\pi\epsilon_0}{\frac{1}{a}-\frac{1}{b}}=\frac{4\pi\epsilon_0~a~b}{b-a}.$$The equivalent capacitance is ##C_{eq}=C_1+C_2.##

The language in the video is incomprehensible to me but not the math. I showed the gist of the argument above for those who don't have the patience to sit through this 4 min video.

This helps to find capacitance without finding induced charges or potential on the spheres.
But there also questions that ask to find charge on inner sphere which causes the problem I asked in the title

Aurelius120 said:
Why though? Why should grouunding the inner sphere cause any problem to the outer sphere? It's still a metal and therefore charge will dustribute to get to maximum distance from each other right?
For the inner sphere to be grounded there must be an insulated hole in the outer conductor. That hole will allow an E field that will disrupt the spherical symmetry near the hole and allow an E field in. So:
Aurelius120 said:
electric field at any point inside a uniformly charged sphereical shell is zero. Thus there is no electrostatic force on the inner sphere.
this doesn’t hold in this problem. There is at least an electrostatic force inside the outer shell near the ground penetration.

Aurelius120 and PeterDonis
Dale said:
For the inner sphere to be grounded there must be an insulated hole in the outer conductor. That hole will allow an E field that will disrupt the spherical symmetry near the hole and allow an E field in. So: this doesn’t hold in this problem. There is at least an electrostatic force inside the outer shell near the ground penetration.
Won't that hole affect the potential (since it affects field)?
Aurelius120 said:
$$\frac{KQ}{b}+\frac{Kq}{a}=0\implies q=\frac{-Qa}{b}$$
Then potential difference$$=\frac{KQ}{b}-\frac{KQa}{b^2}$$
Here the potential of the shells is calculated without accounting for the hole. So value of induced charge on inner sphere is wrong and therefore the rest of the calculations are wrong??

PeterDonis said:
If the inner sphere is grounded, the outer sphere cannot be a uniformly charged spherical shell and Gauss's theorem cannot apply.
If the outer sphere cannot be uniformly charged, the potential difference between them is not equal so they can't be treated as parallel capacitors ?

Aurelius120 said:
Won't that hole affect the potential (since it affects field)?
Yes. That is what grounding is intended to do.

Aurelius120 said:
Here the potential of the shells is calculated without accounting for the hole. So value of induced charge on inner sphere is wrong and therefore the rest of the calculations are wrong??
I think so. The question would be if it is a close approximation. I cannot prove that.

PeterDonis
Aurelius120 said:
If the outer sphere cannot be uniformly charged, the potential difference between them is not equal so they can't be treated as parallel capacitors ?
You might still be able to treat them that way to a good approximation, as @Dale says.

However, you are missing the key point of the hole required for the ground connection: by breaking the continuity of the outer sphere and invalidating Gauss's theorem, it allows a field to be present that accounts for the movement of charge between ground and the inner sphere (through the ground connection) that maintains the potential of the inner sphere at a constant value. So your belief stated in the OP that this change in potential and movement of charge somehow happens without a field driving it is wrong. There is a field driving it.

Aurelius120

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