Why is the net flux through a closed surface equal to zero?

[Frigus]

Suppose we have placed a cube in field which varies linearly with z axis so electric field magnitude on coordinates of face ABCD is clearly more than face EFGH and we know area of both faces are equal,
So if we calculate flux then it would be non zero but it contradicts with the fact that electric field through closed surface enclosing no charge is 0.
Please tell me where I am wrong.

 

[jbriggs444]

Suppose we have placed a cube in field which varies linearly with z axis so electric field magnitude on coordinates of face ABCD is clearly more than face EFGH and we know area of both faces are equal,
So if we calculate flux then it would be non zero but it contradicts with the fact that electric field through closed surface enclosing no charge is 0.
Please tell me where I am wrong.

You have posited a field which requires that a particular charge distribution exists. And then you posited a different charge distribution.

 

[Adesh]

but it contradicts with the fact that electric field through closed surface enclosing no charge is 0.

The fact you’re talking about is applicable only to certain surfaces. You can have a contradiction like this also, consider a surface whose end sides are not parallel but equal in magnitude, and if you place this closed surface in a uniform electric field the net flux will not be zero.

 

[vanhees71]

Your charge density is, according to Gauss's Law
$$\rho=\epsilon_0 \vec{\nabla} \cdot \vec{E}=\rho_0=\text{const}.$$
The flux of the em. field through an arbitrary closed surface $\partial V$ which is the boundary of volume $V$ is
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\int_V \mathrm{d}^3 r \rho(\vec{r})=\frac{V \rho_0}{\epsilon_0}.$$

 

[Ibix]

So if we calculate flux then it would be non zero but it contradicts with the fact that electric field through closed surface enclosing no charge is 0.

As @jbriggs444 says (edit: and @vanhees71), your cube encloses charge. Your electric field is $\vec E=kz\vec{\hat{z}}$, where $k$ is a constant, so $\nabla\cdot\vec E=k$, which Maxwell's equations tells you is the charge density - clearly non-zero.

 

[Frigus]

I got its answer,
Its net flux would be non-zero because this situation is hypothetical as we cannot draw it's field lines because if we diverge or converge(so as to show its magnitude)then its field E would be in y or x direction which Is not possible according to situation,but in real situation electric flux would always be 0 as no. Of field lines entering are always equal to field lines leaving.
Thanks to all of you.

 

[vanhees71]

No! The electric field has a source, electric charges. That's described by Gauß's Law for the electric field. In its fundamental form its written as one of the inhomogeneous local Maxwell equations (in SI units),
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
Using Gauß's integral theorem you can integrate this of a volume $V$ with boundary $\partial V$, leading to
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} \int_{V} \mathrm{d}^3 r \rho=\frac{Q_V}{\epsilon_0}.$$
So if there is charge inside the volume the flux of the electric field through its surface is not 0 but proportional to the total charged contained within the volume.

 

[atyy]

Its net flux would be non-zero because this situation is hypothetical as we cannot draw it's field lines because if we diverge or converge(so as to show its magnitude)then its field E would be in y or x direction which Is not possible according to situation,but in real situation electric flux would always be 0 as no. Of field lines entering are always equal to field lines leaving.

In a situation where the laws of classical electromagnetism hold, the flux enclosed by a closed surface is zero only if the closed surface does not contain any charge. If the closed surface contains charge, the flux enclosed by the surface can be non-zero.

Others have said this above, but it's important, so am repeating the point.

 

[Ibix]

Its net flux would be non-zero because this situation is hypothetical as we cannot draw it's field lines because if we diverge or converge

No. The net flux is non-zero because you have a source of electric field inside your cube.

 

[Frigus]

Can anyone please tell me if this statement "net flux through closed surface is 0 when it is placed in electric field and is enclosing no charge" is true or not.
If it is true then I can't understand why it is?
The explanation I got is that it is because no. Of field lines entering is equal to no. Of field lines leaving but I am unable to understand it if I try to use the mathematical definition of flux i.e ϕ=E.A and I can't find any relation between mathematical and field line definition of flux.
And also we don't have Maxwell equations in our course so I cannot study those as I would have to sacrifice my another chapters to understand it.

 

[A.T.]

Can anyone please tell me if this statement "net flux through closed surface is 0 when it is placed in electric field and is enclosing no charge" is true or not.
If it is true then I can't understand why it is?
The explanation I got is that it is because no. Of field lines entering is equal to no. Of field lines leaving but I am unable to understand ...

If there is no charge enclosed, then no field line begins or ends within the enclosed volume. So any field line that comes in must come out somewhere, and vice versa.

 

[vanhees71]

In your example there WAS a charge inside any volume you can choose and thus the electric flux is as it is. Only if $\vec{\nabla} \cdot \vec{E}=0$ within the volume the flux is 0.

The question "why" about fundamental laws of physics are always hard to answer. Gauß's Law is a fundamental law of nature (as are the Maxwell equations describing the electromagnetic phenomena as far as no quantum effects are relevant, which then is described by the "quantum version", i.e., quantum electrodynamics, aka QED).

So why do Maxwell's equations as they do? You can only such questions, if you define, what you accept as input to answer this question. From a modern point of view, "why questions" should be answered by referring to the fundamental symmetries of nature, and the most important ones to begin with are the symmetries of the spacetime model. For electrodynamics you should always think in terms of relativistic spacetime. Special relativity is enough. So it's Minkowski space with the Poincare group as symmetry group. Take the part that is continuously connected with the identity transformation, it's restricted to the proper orthochronous Poincare group, and now you have to ask about physical laws of motion that obey this symmetry, i.e., any fundamental law must look the same in any inertial reference frame and thus must be invariant under Poincare transformations.

Now electrodynamics should be a Poincare-inveriant local field theory (locality is another assumption, which I can only justify by the pragmatic "because it works"; there's no non-local theory that works in relativistic physics today). Then you can ask which types of fields you have. In the classical domain the most promising idea is to use tensor fields (with tensor referring to the Lorentz group, including of course also scalar and vector fields as special cases). It turns out that the right choice for the electromagnetic field is a massless vector field, but massless vector fields naturally are described as gauge fields (as is derived for the quantum case by asking for the unitary realizations of the proper orthochronous Poincare group, which was first analyzed by Wigner in a famous paper of 1939). Then taking the minimal realization of such a theory for the most simple gauge group U(1) you get almost immediately Maxwell's electromagnetism.

 

[anorlunda]

You can do it graphically, no equations needed. The key is to start with just one point charge.

Above is a case where the point charge is inside in a closed surface. You see the lines all go out. You can redraw the surface to have any shape, and the answer is the same. You could even have a surface where the line goes out, then in again, then out again, then in again, then out again. Subtract ins from outs and the sum for each line is +1.

Below is a case where the point charge is outside of the closed surface. You see that the number of lines going into the surface exactly match the number going out. Outs minus ins equals zero. Try redrawing the surface to any shape that does not enclose the point charge. The answer is the same.

Finally think of any collection of point charges, many points in any spatial distribution. Just add up the lines for each. What applies for one point applies to many.

 

[Frigus]

You can do it graphically, no equations needed. The key is to start with just one point charge.

View attachment 264629

Above is a case where the point charge is inside in a closed surface. You see the lines all go out. You can redraw the surface to have any shape, and the answer is the same. You could even have a surface where the line goes out, then in again, then out again, then in again, then out again. Subtract ins from outs and the sum for each line is +1.

Below is a case where the point charge is outside of the closed surface. You see that the number of lines going into the surface exactly match the number going out. Outs minus ins equals zero. Try redrawing the surface to any shape that does not enclose the point charge. The answer is the same.

View attachment 264628

Finally think of any collection of point charges, many points in any spatial distribution. Just add up the lines for each. What applies for one point applies to many.

It was very satisfying explanation but the only thing I can't understand is that what is relation between this $\vec E \cdot \vec A$ formula and no. Of field lines passing through surface.
This question has took sleeps of my night,please help me in this case also.

 

[Frigus]

In your example there WAS a charge inside any volume you can choose and thus the electric flux is as it is. Only if $\vec{\nabla} \cdot \vec{E}=0$ within the volume the flux is 0.

The question "why" about fundamental laws of physics are always hard to answer. Gauß's Law is a fundamental law of nature (as are the Maxwell equations describing the electromagnetic phenomena as far as no quantum effects are relevant, which then is described by the "quantum version", i.e., quantum electrodynamics, aka QED).

So why do Maxwell's equations as they do? You can only such questions, if you define, what you accept as input to answer this question. From a modern point of view, "why questions" should be answered by referring to the fundamental symmetries of nature, and the most important ones to begin with are the symmetries of the spacetime model. For electrodynamics you should always think in terms of relativistic spacetime. Special relativity is enough. So it's Minkowski space with the Poincare group as symmetry group. Take the part that is continuously connected with the identity transformation, it's restricted to the proper orthochronous Poincare group, and now you have to ask about physical laws of motion that obey this symmetry, i.e., any fundamental law must look the same in any inertial reference frame and thus must be invariant under Poincare transformations.

Now electrodynamics should be a Poincare-inveriant local field theory (locality is another assumption, which I can only justify by the pragmatic "because it works"; there's no non-local theory that works in relativistic physics today). Then you can ask which types of fields you have. In the classical domain the most promising idea is to use tensor fields (with tensor referring to the Lorentz group, including of course also scalar and vector fields as special cases). It turns out that the right choice for the electromagnetic field is a massless vector field, but massless vector fields naturally are described as gauge fields (as is derived for the quantum case by asking for the unitary realizations of the proper orthochronous Poincare group, which was first analyzed by Wigner in a famous paper of 1939). Then taking the minimal realization of such a theory for the most simple gauge group U(1) you get almost immediately Maxwell's electromagnetism.

Thanks for writing a long post and sorry as I can't understand it as this is beyond my level and I am unable to interpret most of the words in this post.

 

[Doc Al]

It was very satisfying explanation but the only thing I can't understand is that what is relation between this $\vec E \cdot \vec A$ formula and no. Of field lines passing through surface.

Perhaps this will help: http://labman.phys.utk.edu/phys222core/modules/m1/Gauss%20law.html

Read the first section on flux, which explains how the number of field lines passing through an area is given by that expression.

 

[Frigus]

Perhaps this will help: http://labman.phys.utk.edu/phys222core/modules/m1/Gauss%20law.html

Read the first section on flux, which explains how the number of field lines passing through an area is given by that expression.

This link is not working

 

[Doc Al]

This link is not working

Very strange! (Sorry about that.)

I'll try to cut and paste the section, see if it works:

Flux
The electric field is a vector field. It is a quantity with magnitude and direction defined at every point in space. Another example of a vector field that is easier to visualize is the velocity of water in a stream. The flux of a vector field through a surface area is the amount of whatever the field represents passing through the area. The total flux depends on strength of the field, the size of the surface area it passes through, and on how the area is oriented with respect to the field. You can think of flux as the amount of something crossing a surface. The surface is a two dimensional (real or imagined) boundary. It can be open or closed. An open surface could be a the area of a door, the area of a sheet of paper, the area of a bowl, etc. A closed surface could be the surface are of a sphere or a cube, etc. Flux is measure at a single point in time. Flux is the total amount of something crossing the surface, it is not something per unit area, etc.

Field lines help us to visualize the electric field. The density of the field lines is proportional to the strength of the field.

The number of field lines passing through a geometrical surface of given area A depends on three factors.

• the strength of the field
• the surface area
• the orientation of the surface

The number of field lines passing through an area A is proportional to the flux through that area.
The diagram on the right shows a locally uniform electric field E. The lines are parallel and have constant density. The same surface is inserted in three different orientations. The maximum number of field lines is intercepted when unit vector normal to the surface, n, is parallel to the field E, while no field lines pass through the surface when n is perpendicular to the field. In general, the number of field lines passing through an area A is directly proportional to A*cosθ, where θ is the angle between the field direction and the unit vector n normal to the surface. This leads to the definition of the electric flux.

ΔΦE = E ΔA cosθ.

ΔΦE is the electric flux through some small area ΔA, whose normal make an angle θ with the direction of the electric field. E is the magnitude of the field. The SI unit of flux is Nm2/C.

E is a vector quantity. It is useful also to represent the area A by a vector A. The length of this vector is the size of the area, while its orientation is perpendicular to the area. It is in the direction of the normal n. We have A = An. The normal to the surface can point into two different directions. For a closed surface, by convention, the normal points outward. With our definition of A we can write the flux as the dot product of E and ΔA.

ΔΦE = E∙ΔA = Eperpendicular*ΔA = E ΔA cosθ.

ΔΦE is the flux through a small are ΔA, which may be part of a larger area A. The total electric flux ΦE through A can be evaluated by summing the differential flux over the all elements of surface A,

ΦE= ∑ΔA -> 0 Eperpendicular ΔA = ∑ΔA -> 0 E∙ΔA.

The flux through a given surface can be positive or negative, since the cosine can be positive or negative. The flux through a closed surface is positive if there is a net outward flow, and negative if there is a net inward flow. We have a net outward flux if there is a source inside the closed surface and a net inward flux if there is a sink inside the closed surface.

Consider the electrical flux passing through a cubical surface with two of its faces perpendicular to a uniform electrical field. The flux passing through the top, bottom, front, and back sides of the cube is zero since these sides are parallel to the field lines and thus do not intercept any of them. The normal vector n is perpendicular to the field for these sides and cosθ is zero. As drawn, the field lines are parallel to the normal vector n for the right side, so the flux through this side is ΦE = EA. The field lines are anti-parallel to the normal vector n for the left side, so the flux through this side is ΦE = -EA. The total flux through the surface of the cube is the sum of the fluxes through all sides, and it is zero.

The flux of a vector field through a closed surface is always zero if there is no source or sink of the vector field in the volume enclosed by the surface. The sources and sinks for the electric field are charges.

The flux of a electric field through a closed surface is always zero if there is no net charge in the volume enclosed by the surface.

 

[rude man]

The fact you’re talking about is applicable only to certain surfaces. You can have a contradiction like this also, consider a surface whose end sides are not parallel but equal in magnitude, and if you place this closed surface in a uniform electric field the net flux will not be zero.

Oh but I think it will.
Remember, flux is the dot product of E times the normal at each point on the closed surface.

 

[A.T.]

This question has took sleeps of my night,please help me in this case also.

This might be helpful:

 

[Frigus]

Very strange! (Sorry about that.)

I'll try to cut and paste the section, see if it works:

Flux
The electric field is a vector field. It is a quantity with magnitude and direction defined at every point in space. Another example of a vector field that is easier to visualize is the velocity of water in a stream. The flux of a vector field through a surface area is the amount of whatever the field represents passing through the area. The total flux depends on strength of the field, the size of the surface area it passes through, and on how the area is oriented with respect to the field. You can think of flux as the amount of something crossing a surface. The surface is a two dimensional (real or imagined) boundary. It can be open or closed. An open surface could be a the area of a door, the area of a sheet of paper, the area of a bowl, etc. A closed surface could be the surface are of a sphere or a cube, etc. Flux is measure at a single point in time. Flux is the total amount of something crossing the surface, it is not something per unit area, etc.

Field lines help us to visualize the electric field. The density of the field lines is proportional to the strength of the field.

View attachment 264633The number of field lines passing through a geometrical surface of given area A depends on three factors.

• the strength of the field
• the surface area
• the orientation of the surface

The number of field lines passing through an area A is proportional to the flux through that area.
The diagram on the right shows a locally uniform electric field E. The lines are parallel and have constant density. The same surface is inserted in three different orientations. The maximum number of field lines is intercepted when unit vector normal to the surface, n, is parallel to the field E, while no field lines pass through the surface when n is perpendicular to the field. In general, the number of field lines passing through an area A is directly proportional to A*cosθ, where θ is the angle between the field direction and the unit vector n normal to the surface. This leads to the definition of the electric flux.

ΔΦE = E ΔA cosθ.

ΔΦE is the electric flux through some small area ΔA, whose normal make an angle θ with the direction of the electric field. E is the magnitude of the field. The SI unit of flux is Nm2/C.

E is a vector quantity. It is useful also to represent the area A by a vector A. The length of this vector is the size of the area, while its orientation is perpendicular to the area. It is in the direction of the normal n. We have A = An. The normal to the surface can point into two different directions. For a closed surface, by convention, the normal points outward. With our definition of A we can write the flux as the dot product of E and ΔA.

ΔΦE = E∙ΔA = Eperpendicular*ΔA = E ΔA cosθ.

ΔΦE is the flux through a small are ΔA, which may be part of a larger area A. The total electric flux ΦE through A can be evaluated by summing the differential flux over the all elements of surface A,

ΦE= ∑ΔA -> 0 Eperpendicular ΔA = ∑ΔA -> 0 E∙ΔA.

View attachment 264634The flux through a given surface can be positive or negative, since the cosine can be positive or negative. The flux through a closed surface is positive if there is a net outward flow, and negative if there is a net inward flow. We have a net outward flux if there is a source inside the closed surface and a net inward flux if there is a sink inside the closed surface.

Consider the electrical flux passing through a cubical surface with two of its faces perpendicular to a uniform electrical field. The flux passing through the top, bottom, front, and back sides of the cube is zero since these sides are parallel to the field lines and thus do not intercept any of them. The normal vector n is perpendicular to the field for these sides and cosθ is zero. As drawn, the field lines are parallel to the normal vector n for the right side, so the flux through this side is ΦE = EA. The field lines are anti-parallel to the normal vector n for the left side, so the flux through this side is ΦE = -EA. The total flux through the surface of the cube is the sum of the fluxes through all sides, and it is zero.

The flux of a vector field through a closed surface is always zero if there is no source or sink of the vector field in the volume enclosed by the surface. The sources and sinks for the electric field are charges.

The flux of a electric field through a closed surface is always zero if there is no net charge in the volume enclosed by the surface.

Thanks,
I get more clear idea what is relation between them.
But their is one another thing which i can't understand.
In the figure I have posted if we try to find flux using formula then we can clearly see that flux of electric field lines leaving is less than that of entering field lines as field lines which are leaving are at some angle(so no.of field lines penterating perpendicularly are less).
We can see that net flux is 0 as no. Of field lines entering are equal to leaving but if we try to find it using formula it seems to me that it should be non-zero.

 

[Doc Al]

We can see that net flux is 0 as no. Of field lines entering are equal to leaving but if we try to find it using formula it seems to me that it should be non-zero.

Your diagram only has three field lines so it's unclear what you are trying to show. If there is no charge in the closed volume, the net flux will be zero. To use the formula you'll need the electric field values.

 

[Ibix]

Of field lines entering are equal to leaving but if we try to find it using formula it seems to me that it should be non-zero.

You shouldn't take "field lines" too literally. They're integral curves of the field, but that means there are infinitely many of them - one passing through every one of the infinite number of points on each side. If you draw a great many of the lines you can count the number passing through each side (dotted with the vector area of the part of the side they pass through, as you note) - but this is just using a sum as an approximation to an integral. That is, $\sum_{i=1}^n \vec{E}_i\cdot\delta\vec{A}_i\approx\int\vec E\cdot d\vec A$ if the number of elements, $n$, in the sum is large. That certainly won't work if you only draw three field lines, though, because the sum is an extremely poor approximation to the integral.

 

[anorlunda]

In the figure I have posted if we try to find flux using formula then we can clearly see that flux of electric field lines leaving is less than that of entering field lines as field lines which are leaving are at some angle

Do you think the number of lines per square meter entering must match the number of lines per square meter leaving? No, that is incorrect. Strike out the "per square meter" phrases and it becomes correct.

 

[A.T.]

In the figure I have posted if we try to find flux using formula then we can clearly see that flux of electric field lines leaving is less than that of entering field lines as field lines which are leaving are at some angle(so no.of field lines penterating perpendicularly are less).

As @anorlunda notes, the number of field lines corresponds to flux itself, not to flux density. Same number of field lines means same flux. Distributing them over a greater area (by spreading them out radially or changing to angle to the surface) represents a reduction flux density.

 

[sophiecentaur]

You shouldn't take "field lines" too literally.

I agree but a very intuitive explanation, which involves field lines goes: How could a situation exist where the number of field lines entering an region is less then the field lines leaving it? There would have to be a 'source' of field lines inside (new lines would need 'ends' on them). In electrostatics, the only source I can think of is a Charge. This is independent of the uniformity or otherwise of flux density.

Next step in this is to believe what the integrals tell you.