# Huge questions about electric fields

#### fluidistic

Gold Member
I can understand that in the middle of an empty conductor sphere the electric field is worth 0. But I don't understand why it is also 0 in the interior of the sphere at any point. I know that by Gauss' law it is 0 but I don't grasp it. For instance if you chose a point very close to the surface, why is it the same as choosing the middle point? Still by Gauss'law?
Well, but then why is the field outside the sphere not 0? If I chose any Gaussian surface outside the sphere, by Gauss' law the electric field should be 0 since there's no charge enclosed. But it is not so. I don't understand why should I enclose the sphere. I could not enclose it if I want not to.
Am I mixing things? Like the net flux passing through any Gaussian surface which is indeed 0 outside the sphere if I chose not to enclose it and it is also 0 inside the sphere. I feel I'm all confused about this.

I don't know why should I enclose or not a charge when using Gauss law.

Another question : Say I have 2 conductor charged spheres. Outside both spheres there exist an electric field, right? But why is the electric field induced by one sphere stopped by the surface of the other sphere? Is it because of Faraday's cage?

#### Jolb

Think of it like this: in a conductor of ANY shape, if there ever WERE an electric field, the free charges of the conductor would just be accelerated into a new configuration. Free charges always move in nonzero fields. Eventually they find the configuration where the field cancels to zero! The charges only accumulate where they're bound (where they're not "free" charge), and they're only bound to staying in the conductor; i.e., they distribute themselves over the surface of the conductor. This is independent of shape.

If you had a charged conducting sphere, the field and the charge configuration would be equivalent to a 2-dimensional spherical shell of that charge. Doing Gauss' law about the center of the sphere, we find that there's zero charged enclosed in the surface if its radius is less than that of the sphere's, and all the charge of the sphere is enclosed once the radius is greater than that of the sphere's.

Edit: you might like to try solving Coulomb's force law for an arbitrary point inside a charged sphere. It does work out to zero if you're patient enough plod through the trig to solve it in general for any point in the sphere. This is equivalent to solving for the gravitational field at an arbitrary point inside a massive hollow sphere.

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#### fluidistic

Gold Member
Think of it like this: in a conductor of ANY shape, if there ever WERE an electric field, the free charges of the conductor would just be accelerated into a new configuration. Free charges always move in nonzero fields. Eventually they find the configuration where the field cancels to zero! The charges only accumulate where they're bound (where they're not "free" charge), and they're only bound to staying in the conductor; i.e., they distribute themselves over the surface of the conductor. This is independent of shape.

If you had a charged conducting sphere, the field and the charge configuration would be equivalent to a 2-dimensional spherical shell of that charge. Doing Gauss' law about the center of the sphere, we find that there's zero charged enclosed in the surface if its radius is less than that of the sphere's, and all the charge of the sphere is enclosed once the radius is greater than that of the sphere's.

Edit: you might like to try solving Coulomb's force law for an arbitrary point inside a sphere. It does work out to zero if you're patient enough plod through the trig to solve it in general for any point in the sphere. This is equivalent to solving for the gravitational field at an arbitrary point inside a massive hollow sphere.
Considering your second paragraph and going where my confusion belongs : If I chose a Gaussian surface that is outside and far away (say 10 times the radius of the sphere or circle if you consider the 2 dimensional problem). It won't enclose any charge, yet the electric field there is not null. Although the net flux passing through the surface is null (I understand this thanks to Gauss' law, but I don't understand why there exist an electric field).

#### Jolb

Considering your second paragraph and going where my confusion belongs : If I chose a Gaussian surface that is outside and far away (say 10 times the radius of the sphere or circle if you consider the 2 dimensional problem). It won't enclose any charge, yet the electric field there is not null. Although the net flux passing through the surface is null (I understand this thanks to Gauss' law, but I don't understand why there exist an electric field).
I'm not really getting you. If you had a conducting sphere of radius A with a charge Q, then a concentric sphere of radius 10A would enclose exactly Q, and so the field at 10A would be proportional to Q(10A)^-2.

If you have a net neutral conducting sphere, the field everywhere is zero.

Any time your gaussian surface encloses no charge, the field on the surface will be equal to zero (provided there's no external charges).

#### fluidistic

Gold Member
I'm not really getting you. If you had a conducting sphere of radius A with a charge Q, then a concentric sphere of radius 10A would enclose exactly Q, and so the field at 10A would be proportional to Q(10A)^-2.

If you have a net neutral conducting sphere, the field everywhere is zero.

Any time your gaussian surface encloses no charge, the field on the surface will be equal to zero (provided there's no external charges).
I wasn't clear enough. Consider a Gaussian surface (take a sphere) outside the charged conductor sphere, such that the Gaussian sphere does not enclose the charged sphere. By Gauss' law there is no net flux passing through the Gaussian sphere, however there is an electric field in the Gaussian sphere although there's no charges.
But now you wrote "(provided there's no external charges).". I think you just made me understand.
If there's a charged sphere or charges and I want to calculate the electric field at ANY point in the space, I MUST chose a Gaussian surface that enclose these charges. Am I right on this?

#### Jolb

I wasn't clear enough. Consider a Gaussian surface (take a sphere) outside the charged conductor sphere, such that the Gaussian sphere does not enclose the charged sphere. By Gauss' law there is no net flux passing through the Gaussian sphere, however there is an electric field in the Gaussian sphere although there's no charges.
But now you wrote "(provided there's no external charges).". I think you just made me understand.
If there's a charged sphere or charges and I want to calculate the electric field at ANY point in the space, I MUST chose a Gaussian surface that enclose these charges. Am I right on this?
The short answer is yes. If you want to use Gauss' law to find the total electric field, you have to enclose all the charges. Gauss' law isn't really the fundamental law for calculating electric field--the fundamental law is Coulomb's force law.

Gauss' law is really a shortcut because it makes use of the divergence theorem to connect the (oriented) flux of a field through a closed surface to the field sources enclosed in the surface. It says little about what's going on outside the surface!

To elaborate, in your example of the gaussian sphere enclosing no charge outside a charged sphere, if you found the field using Coulomb's equation and calculated the net flux through the gaussian sphere (you can do this with trig) you'd get zero. This is generally true for any surface in any conservative field. What this would tell you is that there's no charge enclosed in that sphere/surface.

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#### atyy

Maybe it's not by Gauss's law alone.

If curl E is zero, then Gauss's law can be written as Poisson's equation, which with zero charge becomes Laplace's equation. Then a solution of the equation with an equipotential boundary is that the potential is constant throughout the region. Then use a uniqueness theorem that this must be the only solution. A constant potential means no E field.

#### fluidistic

Gold Member
The short answer is yes. If you want to use Gauss' law to find the total electric field, you have to enclose all the charges. Gauss' law isn't really the fundamental law for calculating electric field--the fundamental law is Coulomb's force law.

Gauss' law is really a shortcut because it makes use of the divergence theorem to connect the (oriented) flux of a field through a closed surface to the field sources enclosed in the surface. It says little about what's going on outside the surface!

To elaborate, in your example of the gaussian sphere enclosing no charge outside a charged sphere, if you found the field using Coulomb's equation and calculated the net flux through the gaussian sphere (you can do this with trig) you'd get zero. This is generally true for any surface in any conservative field. What this would tell you is that there's no charge enclosed in that sphere/surface.
Ok thank you, all is much clearer now to me.

Maybe it's not by Gauss's law alone.

If curl E is zero, then Gauss's law can be written as Poisson's equation, which with zero charge becomes Laplace's equation. Then a solution of the equation with an equipotential boundary is that the potential is constant throughout the region. Then use a uniqueness theorem that this must be the only solution. A constant potential means no E field.
Thanks for the reply. Ok. In my example I'm sure that outside the charged conductor sphere the curl of E is not 0, hence there's an electric field. I appreciate your precision.

Edit: To Jolb : If Gauss' law is not a fundamental law to calculate electric fields but Coulomb's law is, then why do we call Maxwell's equations fundamental laws? By this I mean, why do we say that all the theory of electromagnetism can be derived from Maxwell's equations?

#### Gerenuk

Not sure if any of the following has already been said

I must say Gauss law is as fundamental as Coulombs law. I fact it is mathematically the same.

For instance if you chose a point very close to the surface, why is it the same as choosing the middle point?
There will be many points on the opposite side, that cancel the force from the nearby charges.

Well, but then why is the field outside the sphere not 0? If I chose any Gaussian surface outside the sphere, by Gauss' law the electric field should be 0 since there's no charge enclosed.
The "sum" of the forces on this surface is zero, but not each individual one! Some forces point inside, some point outside on this surface. Only for complete spherical symmetry you can conclude that the force should be constant over the surface (and hence zero if there is no charge inside).

I don't know why should I enclose or not a charge when using Gauss law.
Enclose whenever you wish. But remember that the force over the surface isn't constant in general, so you cannot deduce anything from enclosing.

But why is the electric field induced by one sphere stopped by the surface of the other sphere? Is it because of Faraday's cage?
It is not stopped. It continues forever. However the second plate contributes a field on the other side which exactly cancels the force from the first plate.

#### fluidistic

Gold Member
Not sure if any of the following has already been said

I must say Gauss law is as fundamental as Coulombs law. I fact it is mathematically the same.

There will be many points on the opposite side, that cancel the force from the nearby charges.
Okay!

Gerenuk said:
The "sum" of the forces on this surface is zero, but not each individual one! Some forces point inside, some point outside on this surface. Only for complete spherical symmetry you can conclude that the force should be constant over the surface (and hence zero if there is no charge inside).

Enclose whenever you wish. But remember that the force over the surface isn't constant in general, so you cannot deduce anything from enclosing.
By force, do you mean flux, or electric field value at a point? Or force?

Gerenuk said:
It is not stopped. It continues forever. However the second plate contributes a field on the other side which exactly cancels the force from the first plate.
What? Sorry, I don't understand this part. You have 2 charged conductor spheres. They have the same radius and their charges are worth Q and -Q. How do the electric field induced by the sphere Q is canceled out by the second sphere? What do you mean by plate? A sphere?
How can I prove this mathematically? I'll try to do it, even with 2 spheres with equal charge. But please shred some light on me!

#### Gerenuk

By force, do you mean flux, or electric field value at a point? Or force?
Sorry, I was sloppy about the terms used. Basically Gauss's law is that the total flux equals the inside charge.
$$\iint \vec{E}\cdot\mathrm{d}\vec{S}=\frac{Q}{\varepsilon_0}$$
Do you know what that means?
And only for rotational symmetry of the inside and the environment around the closed surface you can deduce that $\vec{E}\cdot\mathrm{d}\vec{S}[/tex] is constant and therefore in such a case Gauss's law simplifies to [itex]E=\frac{Q}{\varepsilon_0 S}$.

What? Sorry, I don't understand this part. You have 2 charged conductor spheres. They have the same radius and their charges are worth Q and -Q. How do the electric field induced by the sphere Q is canceled out by the second sphere? What do you mean by plate? A sphere?
How can I prove this mathematically? I'll try to do it, even with 2 spheres with equal charge. But please shred some light on me!
Sorry, I confused spheres with plates, but the argument stays the same. The force of a sphere is the same if all the charge were concentrated at the center of the sphere
http://en.wikipedia.org/wiki/Shell_theorem
For two oppositely charged spheres this is equalent to two opposite charge at the common center. These two are just opposite in sign and therefore cancel. So you consider both fields independently and in an infinite region, but outside both sphere these fields exactly cancel.

Does this help? I haven't explained in full detail. Let me know if you need more details.

#### fluidistic

Gold Member
Sorry, I was sloppy about the terms used. Basically Gauss's law is that the total flux equals the inside charge.
$$\iint \vec{E}\cdot\mathrm{d}\vec{S}=\frac{Q}{\varepsilon_0}$$
Do you know what that means?
And only for rotational symmetry of the inside and the environment around the closed surface you can deduce that $\vec{E}\cdot\mathrm{d}\vec{S}[/tex] is constant and therefore in such a case Gauss's law simplifies to [itex]E=\frac{Q}{\varepsilon_0 S}$.
Ok I get it. Yes, I understand Gauss law. That the net flux passing through a closed surface is proportional to the charge the surface encloses. My doubt was that if I chosen a Gaussian surface that does not enclose any charge, although the net flux is null, the electric field is not necessarily null. I understand it now. Because I was used to calculate electric fields of solids using Gauss' law and I didn't know why I should enclose or not charges. Now I know I must enclose them.

Generuk said:
Sorry, I confused spheres with plates, but the argument stays the same. The force of a sphere is the same if all the charge were concentrated at the center of the sphere
http://en.wikipedia.org/wiki/Shell_theorem
For two oppositely charged spheres this is equalent to two opposite charge at the common center. These two are just opposite in sign and therefore cancel. So you consider both fields independently and in an infinite region, but outside both sphere these fields exactly cancel.

Does this help? I haven't explained in full detail. Let me know if you need more details.
I never thought like this before, I like "your" way of seeing the situation. There's something I don't understand well : you say that in the case of the 2 charged conductor spheres, the electric field cancels out outside the spheres. However if I consider their fields independently, I realize that it depends on R² where R is the distance of the center of the sphere (as you said, as if its charge was concentrated into its center). So I'd have an electric field of a dipole. It is not null as far as I know. Did I misunderstood you on this? I understood that it's like having 2 charges at the same point, precisely in the center of mass of the 2 spheres (if they are equal in radius, but with opposed charge), but the 2 charges cancels out since it's Q-Q. But then, where's the dipole?

My doubt was not this. It is that inside a sphere A (vacuum), the electric field is null, even if there's a similar sphere B close to the sphere A. Like if the electric field of the sphere B cannot enter through the surface of sphere A. Why is it so? Is it because of a Faraday's cage effect?

#### atyy

Thanks for the reply. Ok. In my example I'm sure that outside the charged conductor sphere the curl of E is not 0, hence there's an electric field. I appreciate your precision.
In electrostatics, curl E is also zero outside the conductor. The difference is that at infinity, the potential must go to zero, which is a different boundary condition for Laplace's equation. So the potential cannot be constant, and there's an electric field outside the conductor.

#### fluidistic

Gold Member
In electrostatics, curl E is also zero outside the conductor. The difference is that at infinity, the potential must go to zero, which is a different boundary condition for Laplace's equation. So the potential cannot be constant, and there's an electric field outside the conductor.
Wow, thanks. I'm starting the real (1 year long) EM course in March. I'm sure I'll understand things better than I do right now, despite the fact that I've taken calculus 3.
So I'll likely read this post again when I'll have a better understanding.

#### Gerenuk

However if I consider their fields independently, I realize that it depends on R² where R is the distance of the center of the sphere (as you said, as if its charge was concentrated into its center). So I'd have an electric field of a dipole. It is not null as far as I know. Did I misunderstood you on this? I understood that it's like having 2 charges at the same point, precisely in the center of mass of the 2 spheres (if they are equal in radius, but with opposed charge), but the 2 charges cancels out since it's Q-Q. But then, where's the dipole?
In general the dipole contribution is
$$\vec{E}_\text{dipole}=\iiint \vec{r}\mathrm{d}\varrho$$
Anyway, in this case the charges are exactly at the same point. So there are no higher order contributions.
A further note: For any configuration of charges you can write the total electric field either with the individual components
$$\vec{E}(\vec{r})=\sum\frac{\alpha_i}{|\vec{r}_i-\vec{r}|^2}$$
OR the multipole contributions from this configuration (which is a mathematical transformation)
$$\vec{E}=\frac{\sum \alpha_i}{|\vec{r}|^2}+\frac{\sum \alpha_i\vec{r_i}}{|\vec{r}|^3}+\text{(quadrupole)}+\dotsb$$
Note the isolated appearance of $\vec{r}$ and $\vec{r}_i$ here.

My doubt was not this. It is that inside a sphere A (vacuum), the electric field is null, even if there's a similar sphere B close to the sphere A. Like if the electric field of the sphere B cannot enter through the surface of sphere A. Why is it so? Is it because of a Faraday's cage effect?
The cage effect is also a cancellation. Mobile charges on the surface of a conductor will rearrange to create a field that exactly cancels the outside field.
In your case there a static charges. Fields penetrate all matter and do not stop, however they also add in superposition. And in your case with the spheres, the fields of all elements of the spheres summed together again cancel out exactly.

#### fluidistic

Gold Member
For two oppositely charged spheres this is equalent to two opposite charge at the common center. These two are just opposite in sign and therefore cancel. So you consider both fields independently and in an infinite region, but outside both sphere these fields exactly cancel.
,

In general the dipole contribution is
$$\vec{E}_\text{dipole}=\iiint \vec{r}\mathrm{d}\varrho$$
Anyway, in this case the charges are exactly at the same point. So there are no higher order contributions.
If instead of the 2 spheres we chose an electron with a proton (the common hydrogen atom), does that mean that there's no electric field at atomic scale?

Generuk said:
The cage effect is also a cancellation. Mobile charges on the surface of a conductor will rearrange to create a field that exactly cancels the outside field.
In your case there a static charges. Fields penetrate all matter and do not stop, however they also add in superposition. And in your case with the spheres, the fields of all elements of the spheres summed together again cancel out exactly.
I think I get this, thanks a million!

#### Gerenuk

If instead of the 2 spheres we chose an electron with a proton (the common hydrogen atom), does that mean that there's no electric field at atomic scale?
If those two sphere real were displaced, then you indeed would have a dipole field. For perfectly concentric spheres there is no dipole field.

#### fluidistic

Gold Member
If those two sphere real were displaced, then you indeed would have a dipole field. For perfectly concentric spheres there is no dipole field.
I wasn't aware of this. By displaced, do you mean in motion? For example, in the impossible case of a proton and an electron which are not moving, there wouldn't be any electric field? But if they are moving yes?

#### diazona

Homework Helper
If you had a proton and an electron which were not moving (and this is perfectly possible in classical electromagnetism), there would be an electric field. However, if they could somehow exist in exactly the same location, their electric fields would cancel out and there would be no net electric field. You can't do that with a proton and electron because they're solid particles (well, we pretend they are anyway), but you can do it with spheres: if you place the two spheres so that their centers are in the same place, there is no electric field outside the spheres.

#### fluidistic

Gold Member
If you had a proton and an electron which were not moving (and this is perfectly possible in classical electromagnetism), there would be an electric field. However, if they could somehow exist in exactly the same location, their electric fields would cancel out and there would be no net electric field. You can't do that with a proton and electron because they're solid particles (well, we pretend they are anyway), but you can do it with spheres: if you place the two spheres so that their centers are in the same place, there is no electric field outside the spheres.
Ah ok, I see what you mean now... Thanks.

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