Another way of stating Gauss' law?

[Delta2]

How is that possible? If there is now a resultant field, if the field is changed now then the net flux must change. What kind of law is this?

That's how Gauss's law goes whether you like it or not. Changes in the resultant field, result to changes in the net flux through a closed surface only if those changes in the resultant field are due to changes in the charges enclosed by the closed surface.

 

[vanhees71]

It's a mathematical theorem (see my posting above).

 

[Delta2]

It's a mathematical theorem (see my posting above).

Sorry I don't think this has anything to do with the divergence theorem. The divergence theorem is used to prove the equivalence of Gauss's law in integral and differential form.

 

[rudransh verma]

That's how Gauss's law goes whether you like it or not. Changes in the resultant field, result to changes in the net flux through a closed surface only if those changes in the resultant field are due to changes in the charges enclosed by the closed surface.

But you said the flux changes locally ie of one side. What about that?

 

[vanhees71]

Well yes, I tacitly assumed we take Maxwell's equations for granted. I hope we agree on this?

Then it's just Gauss's integral theorem that
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} \int_V \mathrm{d}^3 r \rho=\frac{1}{\epsilon_0} Q_V.$$

 

[vanhees71]

But you said the flux changes locally ie of one side. What about that?

What do you mean by "one side"? The flux in Gauss's law is always through a closed (!) surface!

 

[Delta2]

But you said the flux changes locally ie of one side. What about that?

Yes it can change in one side (for example upper half) but it won't change totally (through the whole closed surface), that is upper half+lower half, if the enclosed charge doesn't change. That is simply Gauss's law, changes in flux through a closed surface occur only when the enclosed charge changes. But changes in flux through an open surface can occur even if the enclosed charge doesn't change. Gauss's law is only for the net flux through closed surfaces.

 

[rudransh verma]

You are saying the resultant field will change near the surface but the net flux will not because it only depends on charge inside the surface which is constant and outside charge contribute to zero flux. The flux x is the flux of the field that is coming from inside charge which will not change because charge doesn’t change. Inside charge is same. Only the resultant field is changing. The field from the inside charge doesn’t. So the flux x is not changing. It’s the same. So net phi= x+y-y=x.
We only talk of the flux of that field that is originating from a charge. There is still the same field from inside charge which can’t be altered. So it means there are three field vectors . One from outside charge, one from inside charge and the resultant.

@Delta2 Let me make this more clear. You are saying the outside field from charge Q cannot alter the field inside the surface from enclosed charge. It can only add up with the charge q’s field only outside the surface. That’s where the flux changes locally. But the net flux remains same because the inside field is same as before being pierced by the charge Q. The field of charge Q enter and leave the surface. So phi= y-y+x=x. x is the flux of field inside the surface that never changes because enclosed charge never changes.

 

[rudransh verma]

@Delta2 I am unable to picture it properly.

 

[rudransh verma]

Yes, but the new field has no sources inside your volume. So its flux through the closed (!) surface must be 0.

Maybe now I got it. The new field has no lines of force. It’s just an effect of two fields due to charge q and Q, Q being outside the surface. There is no lines of force actually penetrating the surface from this new field. The only field of lines are from q andQ.There is no source charge inside the surface. Only source are q and Q combined. And so zero flux due to this field. And so the only fluxes are x, -y, +y which comes x as net flux. So Q charge is useless in gauss law.

 

[Delta2]

@Delta2 I am unable to picture it properly.

Sorry I just can't understand what exactly you don't understand. Gauss's law is for closed surfaces. For closed surfaces the total flux doesn't change (even if the resultant field changes due to an external charge) unless the enclosed charge changes, that's what Gauss's law tell us. For open surfaces, Gauss's law doesn't hold, so the flux over the open surface can change if the resultant field changes.

 

[rudransh verma]

Sorry I just can't understand what exactly you don't understand. Gauss's law is for closed surfaces. For closed surfaces the total flux doesn't change (even if the resultant field changes due to an external charge) unless the enclosed charge changes, that's what Gauss's law tell us. For open surfaces, Gauss's law doesn't hold, so the flux over the open surface can change if the resultant field changes.

Please read my post #45.

 

[Delta2]

Please read my post #45.

Well I read it and I don't understand why you want the resultant field to be some sort of inactive or invisible (without lines of force). Whether you speak of the resultant field, or the sum of the fields from q and Q, it is the same and one thing

 

[rudransh verma]

Well I read it and I don't understand why you want the resultant field to be some sort of inactive or invisible (without lines of force). Whether you speak of the resultant field, or the sum of the fields from q and Q, it is the same and one thing

Because only from a source can come out the field of lines. Like from q and Q. I didn’t say that it is inactive but it is the effect of two fields . Because of the presence of two fields this new field arises but not like there are some new lines of forces.
If we place one charge it will have lines of forces. A test charge will move in some direction. If we place another charge it will bend its line of forces and the direction of test charge also changes. So now there is new effect of the two fields.
Maybe there are new lines of forces but those lines don’t have a source inside the surface penetrating the surface.

 

[Delta2]

Well the resultant field has field lines that are the vector sum of the field lines of the two separate fields.

If we place another charge it will bend its line of forces

Yes the new bended line of forces are the force lines of the resultant field, where is the problem with this view?

 

[rudransh verma]

Well the resultant field has field lines that are the vector sum of the field lines of the two separate fields.

Yes the new bended line of forces are the force lines of the resultant field, where is the problem with this view?

Problem is that the new field has now new flux.

 

[Delta2]

Problem is that the new field has now new flux.

No there is no new flux, either you take the sum of fluxes of the two fields, or the flux of the resultant field, you can't take both.

 

[rudransh verma]

No there is no new flux, either you take the sum of fluxes of the two fields, or the flux of the resultant field, you can't take both.

Let me provide you one reference where we can discuss on common ground. Now focus on surface S1. We place a Q near S1. It will bend its line of forces. There is new field.
you are saying either we take flux x-y+y=x all individual fluxes or flux of resultant field which will be x. Charge outside has no effect on the flux of resultant field. It will be x no matter what charge is outside.

 

[Delta2]

Yes this is all about what we 've been talking about for so many posts. The flux through the surface S1, seems that it must be different since we have now bended lines of force, however it isn't different because that's what Gauss's law tell us, that the flux depends only on the enclosed charge, it does not depend if the field lines are bended, or there is some external charge that bends the field lines , or if the resultant field changes or e.t.c

 

[Delta2]

If I have understood well, I think the cause of your misunderstanding is that you think that the resultant field is a new field that coexists with the two other fields. It depends how you see it but either you ll say that only the resultant field exists, or only the two other fields exist and we ll take their sum when computing the flux. You just can't have all three together.

 

[rudransh verma]

If I have understood well, I think the cause of your misunderstanding is that you think that the resultant field is a new field that coexists with the two other fields. It depends how you see it but either you ll say that only the resultant field exists, or only the two other fields exist and we ll take their sum when computing the flux. You just can't have all three together.

If I say that now there is a new resultant field what will be the flux of S1. x?

 

[Delta2]

Depends what you mean by "new" resultant field. Do you mean that we bring in a 3rd charge in the area? It depends where we do put this new charge on whether the flux through S1 will change or not.

 

[rudransh verma]

Depends what you mean by "new" resultant field. Do you mean that we bring in a 3rd charge in the area? It depends where we do put this new charge on whether the flux through S1 will change or not.

No I meant the resultant of two charges q and Q. If now there is a resultant then still the flux of S1 will be x

 

[Delta2]

No I meant the resultant of two charges q and Q. If now there is a resultant then still the flux of S1 will be x

Yes the flux will still be x.

 

[rudransh verma]

Yes the flux will still be x.

Let me tell: the field of q does bend due to presence of Q. There is a new resultant field but that does not mean it will alter the field of q. It only gets bent ,not changed by the other field. And so the flux of S1 remains same. x.

 

[Ibix]

Let me tell: the field of q does bend due to presence of Q. There is a new resultant field but that does not mean it will alter the field of q. It only gets bent ,not changed by the other field. And so the flux of S1 remains same. x.

The better way to say this is that the resultant electric field $\vec{E}(\vec{r})$ can be written as the sum of two fields $\vec{E}_q(\vec{r})$ and $\vec{E}_Q(\vec{r})$ which are the fields that each charge would have if it were in its present position but the other charge were removed. The total flux of $\vec{E}(\vec{r})$ through $S_1$ (which encloses the charge $q$) is $x$, the same as the total flux of $\vec{E}_q(\vec{r})$ would be if the charge $Q$ were not present.

Note that the flux of $\vec{E}(\vec{r})$ through a small part of $S_1$ will be different from the flux of $\vec{E}_q(\vec{r})$ through the same region - it is only the total over the entire closed surface that is equal.

 

[Delta2]

Yes , well I don't think it can be stated any better than post #61 puts it. @rudransh verma If I try to follow your line of logic I get confused too...

 

[rudransh verma]

@Delta2 i can understand the post #61. But that is not very detailed.
I am asking why about post #61. I want to get an intuitive idea. That is what I am trying to say in 60

 

[Delta2]

@Delta2 i can understand the post #61. But that is not very detailed.
I am asking why about post #61. I want to get an intuitive idea. That is what I am trying to say in 60

Well trying once again to follow your logic:
The new resultant field is the field of q, bended by the field of Q. We don't have three separate fields (unless you view it purely mathematically) we have two fields and the resultant field is their sum. The sum is some sort of bended version of the field of q (or the field of Q) but as I said before many times, either you consider the flux of the sum, or the sum of the fluxes of the field from q and of the field of Q. Mathematically it is $$\Phi_{total}=\iint_{S_1} \vec{E_{total}}\cdot d\vec{A}=\iint_{S_1}\vec{E_q}\cdot d\vec{A}+\iint_{S_1}\vec{E_Q}\cdot d\vec{A}=\Phi_q+\Phi_Q$$

Now it is $\Phi_q=\frac{q}{\epsilon_0}$ but $\Phi_Q=0$. Now if you ask me why it is that $\Phi_Q=0$ though the $E_Q\neq 0$ this is a question I can answer but we will need divergence theorem from vector calculus.

 

[Ibix]

Are you asking why the field lines are curved?

 

[rudransh verma]

The sum is some sort of bended version of the field of q (or the field of Q)

I say if we place a test charge in this resultant field it will move in it according to it. It will not move according to field of q or Q. But this doesn’t mean fields of q and Q are not there physically. Field of Q cannot alter/change/cross the field of q. It cannot manipulate it by its presence. And that’s why even in the presence of Q the field of q is undisturbed and so is the flux x. Bent field is not a changed field.

 

[Delta2]

It cannot manipulate it by its presence. And that’s why even in the presence of Q the field of is undisturbed and so is the flux x.

Yes the field of Q cannot change the field of q. But there going to be a new field that is the sum of Q and q. The flux of this new field , over the surface S1, is the same as the flux of the field of q , over surface S1, but the reason is not that the field from Q cannot manipulate the field of q. The reason is because that's what Gauss's law essentially tell us. The field from Q has zero flux over the surface S1, because Q is not enclosed in S1.

 

[rudransh verma]

The flux of this new field , over the surface S1, is the same as the flux of the field of q , over surface S1, but the reason is not that the field from Q cannot manipulate the field of q. The reason is because that's what Gauss's law essentially tell us. The field from Q has zero flux over the surface S1, because Q is not enclosed

First of all the field lines from Q cannot enter the surface of S1. It gets bent by the presence of q’s field in the way. So even though there is a resultant field there is no effect on the number of lines pierced (flux) of the field of q. Flux remains same.

 

[Delta2]

First of all the field lines from Q cannot enter the surface of S1. It gets bent by the presence of q’s field in the way. So even though there is a resultant field there is no effect on the number of lines pierced (flux) of the field of q. Flux remains same.

No I do not agree, the field from q can't alter the field from Q (neither vice versa). It is the resultant field (their sum) that it is bended. The field from Q enters from one portion of S1 and exits from the back portion of S1, that's why (at least intuitively) the total flux of the field from Q through S1 is zero.

 

[rudransh verma]

The field from Q enters from one portion of S1 and exits from the back portion of S1,

When you place two positive charges near each other what pattern of field lines form. Both the field lines from the charges get bent in front of each other. The field lines from one charge does not cross through the other charge. It gets deviated in the way.