# Another way of stating Gauss' law?

Gold Member
Gauss law relates the net flux phi of an electric field through a closed surface to the net charge q that is enclosed by that surface. It tells us that
Phi = q/permittivity
Can I say it like this : The gauss law states that the net flux of the surface depends upon the net charge enclosed by that surface and it does not depend upon the charge outside the surface.
The charge outside the surface would change the pattern of field but the net flux would not change because it is outside the surface.
So the Q charge outside would not enter the eqn of gauss law in any way.
It feels confusing to me.

Delta2 and vanhees71

I think it's a good way to state Gauss' law. As for the charge outside the closed surface, it's true. It doesn't contribute to the flux.

vanhees71
Gold Member
I think it's a good way to state Gauss' law. As for the charge outside the closed surface, it's true. It doesn't contribute to the flux.
The enclosed charge has effect on net flux of the surface. That is gauss law. But the charge outside has no contribution to net flux. So we don’t write a law like phi=(q+Q)/permittivity.
I think we can insert a charge Q in the law but that will be of no use because the purpose of gauss law is to find Flux and E from net charge and vice versa. It is to find what is inside a box by looking at the box. Q is of no use here. It will only mess up the law. If Q is in the eqn we can never find total charge from the given flux. Also we cannot find field like this.

Last edited:
Hamiltonian299792458
Gold Member
The enclosed charge has effect on net flux of the surface. That is gauss law. But the charge outside has no contribution to net flux. So we don’t write a law like phi=(q+Q)/permittivity.
I think we can insert a charge Q in the law but that will be of no use because the purpose of gauss law is to find Flux and E from net charge and vice versa. It is to find what is inside a box by looking at the box. Q is of no use here. It will only mess up the law. If Q is in the eqn we can never find total charge from the given flux. Also we cannot find field like this.
@Delta2 can you verify this?

Delta2
Homework Helper
Gold Member
@Delta2 can you verify this?
Well , tbh I don't know what would happen if Gauss's law was including the charge outside of the surface, I think one consequence would be that the Coulomb force wouldn't follow an inverse square law. And yes , as you say the whole point is to be able to calculate what's inside the box by looking at the box, we would lose this ability too.

Gold Member
think one consequence would be that the Coulomb force wouldn't follow an inverse square law.
How?
e0E4pir^2=q+Q
E= 1/(4pie0)(q+Q)/r^2
Then?
I don’t follow!

Delta2
Homework Helper
Gold Member
How?
e0E4pir^2=q+Q
E= 1/(4pie0)(q+Q)/r^2
Then?
I don’t follow!
Well, nvm, it seems I was wrong on that conclusion... I cant think of how you would define Q in that case, Q as just the charge outside, might as well been the charge of the whole universe.

vanhees71
Gold Member
The most simple and fundamental form of the electromagnetic laws are the local (differential) Maxwell equations. So it's always most simple to start from them. Gauss's Law for the electric fields read (using SI units)
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
Now use Gauss's integral theorem and integrate this equation over some volume ##V## with its boundary surface ##\partial V##. This leads to
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} \int_{V} \mathrm{d}^3 r \rho=\frac{1}{\epsilon_0} Q_V.$$
This tells you that the flux of the electric field through a closed (!!!) surface is the charge ##Q_V## (times the ugly factor ##1/\epsilon_0## which is due to the choice of SI units) contained in the volume enclosed by this surface.

Demystifier and Delta2
Gold Member
Well, nvm, it seems I was wrong on that conclusion... I cant think of how you would define Q in that case, Q as just the charge outside, might as well been the charge of the whole universe.
This Q charge is enormous charge placed near the surface. Tell me does this field of charge Q only bend the fields of the charge inside the surface. I don’t think it amplifies/cut/intersect that field by interacting with it in any way. Because then the flux/density of the field of inside charge would change by an outside charge Q. It would then enter in gauss law.

Delta2
Homework Helper
Gold Member
This Q charge is enormous charge placed near the surface. Tell me does this field of charge Q only bend the fields of the charge inside the surface. I don’t think it amplifies/cut/intersect that field by interacting with it in any way. Because then the flux/density of the field of inside charge would change by an outside charge Q. It would then enter in gauss law.
In physics we have a principle called the superposition principle. In the case of electric field, when we have two or more sources of electric field, then the field from each source, doesn't alter or interact the field from the other sources, but it just adds up, so we have a total field that is the sum of the fields of each source.

Gold Member
In physics we have a principle called the superposition principle. In the case of electric field, when we have two or more sources of electric field, then the field from each source, doesn't alter or interact the field from the other sources, but it just adds up, so we have a total field that is the sum of the fields of each source.
I know. There is a resultant field with magnitude and direction. Book says the pattern of field changes. Doesn’t this effect the net flux of the surface?

Delta2
Homework Helper
Gold Member
I know. There is a resultant field with magnitude and direction. Book says the pattern of field changes. Doesn’t this effect the net flux?
Yes, the pattern of the field in the surface changes, however this does not changes the net flux through the closed surface because whatever new flux enters our surface due to the outside charge Q, the same flux exits our surface through another segment of the surface, so the net flux =x+y-y=x+0=x where x the flux due to the inside charge and y the flux due to the outside charge.

Hamiltonian299792458
Gold Member
Yes, the pattern of the field in the surface changes, however this does not changes the net flux through the closed surface because whatever new flux enters our surface due to the outside charge Q, the same flux exits our surface through another segment of the surface, so the net flux =x+y-y=x+0=x where x the flux due to the inside charge and y the flux due to the outside charge.
But the flux x will change because the field of inside charge changes. Remember the resultant field. Now there is a new field with new magnitude and direction.

vanhees71
Gold Member
Yes, but the new field has no sources inside your volume. So its flux through the closed (!!!) surface must be 0.

Delta2
Delta2
Homework Helper
Gold Member
But the flux x will change because the field of inside charge changes. Remember the resultant field. Now there is a new field with new magnitude and direction.
No the flux x (which is the flux due to the field of the enclosed charge) does not change, because the field of the enclosed charge does not change. The resultant field at every point at our closed surface will be different, however the resultant net flux (if I can call it that way), will not change, because the new source is not enclosed by our closed surface.

Why this happens? I tried to argued in post #12 in terms of flux that enters and exits, but in fact there is no more why here, it is just what Gauss's law tell us, because Gauss's law tell us that when we calculate the net flux, it will be the same as long as the enclosed charge is the same, regardless of what happens to the resultant field.

Last edited:
Delta2
Homework Helper
Gold Member
Sorry I don't see anything wrong in my grammar at that post. I edited it though to put some paragraph formatting and some additional argument.

Gold Member
No the flux x (which is the flux due to the field of the enclosed charge) does not change, because the field of the enclosed charge does not change. The resultant field at every point at our closed surface will be different, however the resultant net flux (if I can call it that way), will not change, because the new source is not enclosed by our closed surface.

Why this happens? I tried to argued in post #12 in terms of flux that enters and exits, but in fact there is no more why here, it is just what Gauss's law tell us, because Gauss's law tell us that when we calculate the net flux, it will be the same as long as the enclosed charge is the same, regardless of what happens to the resultant field.
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
Delta2
Homework Helper
Gold Member
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.
I agree to all of the above, seems to me you got it perfectly right!

rudransh verma
Gold Member
I agree to all of the above, seems to me you got it perfectly right!
It means the density of the field does not change if there is another field present. A test charge will move in the direction of resultant but the density of individual charge’s field stays same.

Delta2
Homework Helper
Gold Member
It means the density of the field does not change if there is another field present. A test charge will move in the direction of resultant but the density of individual charge’s field stays same.
How do you define the term "density" of a field?

Gold Member
I mean volume of field by density

Delta2
Homework Helper
Gold Member
Ok I see, well, flux changes locally, because the resultant field changes locally, however the total net flux through a closed surface doesn't change if the additional source charges that cause the change of the resultant field are not within our closed surface.

Gold Member
@Delta2 Can you summarise but don’t leave any detail. If one flux changes due to presence of another field then the net flux also change.

Delta2
Homework Helper
Gold Member
@Delta2 Can you summarise but don’t leave any detail. If one flux changes due to presence of another field then the net flux also change.
OK I see now what you meant in post #19 and it is correct.

I understood something else: Suppose we have a charge q, and consider a gaussian spherical surface that surrounds q. Then the net flux is ##\frac{q}{\epsilon_0}##. Now suppose we put another charge q' outside our spherical surface, outside the upper half of the surface. Then the total flux through the surface again remains the same ##\frac{q}{\epsilon_0}##. However if I ask you for the flux of the upper half of our spherical surface, will it be the same or not?

Gold Member
OK I see now what you meant in post #19 and it is correct.

I understood something else: Suppose we have a charge q, and consider a gaussian spherical surface that surrounds q. Then the net flux is ##\frac{q}{\epsilon_0}##. Now suppose we put another charge q' outside our spherical surface, outside the upper half of the surface. Then the total flux through the surface again remains the same ##\frac{q}{\epsilon_0}##. However if I ask you for the flux of the upper half of our spherical surface, will it be the same or not?
I think: the flux of upper half will change because of the changed field. So the net flux will change. I am not clear how two fields behave in each other’s vicinity