I Another way of stating Gauss' law?

[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.

 

[Delta2]

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.

No it is the resultant field that has bended field lines. The constituents of the resultant field each remain as if they were the only field present in space.

I suggest you read about the superposition principle, maybe your book has a section on it, if not try wikipedia
https://en.wikipedia.org/wiki/Superposition_principle

 

[rudransh verma]

No it is the resultant field that has bended field lines. The constituents of the resultant field each remain as if they were the only field present in space.

Here in this figure I can see there are two types of lines. One from one charge and another from other charge. If I take one charge away then the field lines will straighten up. There is no resultant field of lines. We only see resultant‘s effect when we place a test charge. That is what I was trying to say before. There are two constituent fields bent like that but no third resultant field.

 

[Delta2]

No I insist those field lines are the field lines of the resultant field.

 

[vanhees71]

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.

Any field has field lines. If there are no sources inside a closed surface as many field lines enter this surface as leave it and thus the total flux through the closed surface is 0. In electrostatics the charges are sinks and sources of the field lines.

The great thing is that you only need to understand this for a point charge, i.e., the Coulomb fields. All other fields then can be calculated by superposition:
$$\vec{E}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(\vec{x}')(\vec{x}-\vec{x}')}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|^3}.$$

 

[rudransh verma]

No I insist those field lines are the field lines of the resultant field.

I mean there are two fields physically. There is no resultant field physically. Its only an effect.

 

[Delta2]

No I don't agree with this. Of course there is the resultant field physically. You said it yourself many times, if we put a test charge it will move in the direction of the resultant field.

 

[rudransh verma]

No I don't agree with this. Of course there is the resultant field physically. You said it yourself many times, if we put a test charge it will move in the direction of the resultant field.

Yes, it will. Let me explain. If suppose I place a charge it will setup its field all straight up. Now place a test charge. It will start to move in a straight line. Now as it moves a little bit place a second charge in vicinity. What will happen. The path of test charge will deviate. Meaning now the second charge’s field has bent the field of first one. There is no resultant field here. Only an effect.

Suppose there are two forces applied to a body one is horizontal and one is making theta angle with it. Now the body doesn’t move in horizontal direction but in a direction between the two force’s direction. That doesn’t mean there is a resultant force physically. There are only two forces physically, one is horizontal and other making theta angle with it. It’s because there are two forces in two directions that the body now moves in new direction. We say that body moves in the direction of resultant force.

 

[Delta2]

I sense there is some sort of weird misunderstanding in your mind, regarding what exactly a resultant field is but I can't figure out what it is.
I 'll just repeat myself, the resultant field is physical. I got no clue what you mean by "There is no resultant field. Only an effect."

 

[rudransh verma]

I sense there is some sort of weird misunderstanding in your mind, regarding what exactly a resultant field is but I can't figure out what it is.
I 'll just repeat myself, the resultant field is physical. I got no clue what you mean by "There is no resultant field. Only an effect."

What do you think about force. Is there actually a resultant force?

 

[Delta2]

What do you think about force. Is there actually a resultant force?

Depends how you see it. It is the force by which you can replace the constituent forces and it will have the same motional effect as the combination of the constituent forces.

 

[rudransh verma]

Depends how you see it. It is the force by which you can replace the constituent forces and it will have the same motional effect as the combination of the constituent forces.

Yes, it’s either the constituent forces or a resultant force but not both at one time.

 

[Delta2]

Yes, it’s either the constituent forces or a resultant force but not both at one time.

yes that's correct.

 

[rudransh verma]

yes that's correct.

So in the same way it’s either two fields or one resultant field at a time but not both. Effect will be same in either case. And here it’s two fields whose effect is same as if there were one resultant field.

 

[vanhees71]

There is one and only one electromagnetic field. I really don't understand anymore, what this discussion is about :-(.