Problem with electric flux

  • #1
rudransh verma
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Gauss law relates between E at some point on guassian surface with the net charge enclosed by that surface. Using gauss law is like being able to tell what (ie charge)is inside a gift box by just looking at the wrapper(electric field). There are two types of problem. Sometimes we know the charge and want to find field by gauss law. Sometimes we know the field and want to find charge enclosed by the surface.

But we cannot find it all by just looking at the field lines. We need a quantitative way of determining how much field pierces a surface. That measure is called electric flux.
Let’s take a square plate of area A.we define flux at a patch: delta phi= Ecos theta delta A. we define total flux as it’s integration.

But we know that Ecos theta is just one vector at a point. How can multiplying it with area gives us a measure of field of the patch. It would mean that the patch is filled fully with such vectors with no spacing. We know from the patterns of seeds in oil that there are gaps in between the field lines. Where is that that tells as about gaps.

simply integrating each patch would give us a volume of fully filled vectors with no spacing. Where is the gap. Are all points filled with vectors. Surely they would not be same at all points. Or are we talking about uniform field whose value is same at each point. Even then there is nothing that says about gaps in between field lines.
 

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  • #2
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Have you come across how to construct surface integrals? Given some parametric representation ##\mathbf{r} = \mathbf{r}(u,v)## of a surface ##\mathcal{S}##, you have the vector elementary area ##d\mathbf{S} = ( \mathbf{r}_u \times \mathbf{r}_v ) du dv##. If the electric field ##\mathbf{E}(\mathbf{r})## is some function of position then its flux through an infinitesimal portion of ##\mathcal{S}## is ##\mathbf{E} \cdot d\mathbf{S}##, and evaluating the multiple integral over ##u## and ##v## yields the total flux.
 
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  • #3
Delta2
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There are no gaps between field lines, the electric field is defined in every point in space. We just cant draw every possible field line, we draw some of them, that's why you see gaps, but there arent really any gaps.
 
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  • #4
rudransh verma
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There are no gaps between field lines, the electric field is defined in every point in space. We just cant draw every possible field line, we draw some of them, that's why you see gaps, but there arent really any gaps.
So let me tell what I understood. Multiplying E vector with dA gives the volume ie how much field pierces the patch. This volume of field is like flow of water. Here I assume the plate is placed in uniform field. That means field vector at each point on plate is same. Integration of all patches gives total volume of field passing the plate. This field is coming from some source. A field flowing from the plate is like water flowing from a pipe. There are no gaps inside field.
 
  • #5
Delta2
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So let me tell what I understood. Multiplying E vector with dA gives the volume ie how much field pierces the patch. This volume of field is like flow of water. Here I assume the plate is placed in uniform field. That means field vector at each point on plate is same. Integration of all patches gives total volume of field passing the plate. This field is coming from some source. A field flowing from the plate is like water flowing from a pipe. There are no gaps inside field.
I am not sure of what you mean by "volume of field" and of this generic parallelism between electrodynamics and fluid dynamics, but I guess you can view it like this. There are no gaps in the electric field, pretty much like there are no gaps in the flow of a fluid coming from a pipe or another source. The gaps in the drawing of field lines exist because we just can't draw all the field lines, passing from every possible point of the paper.
 
  • #6
rudransh verma
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I am not sure of what you mean by "volume of field" and of this generic parallelism between electrodynamics and fluid dynamics, but I guess you can view it like this. There are no gaps in the electric field, pretty much like there are no gaps in the flow of a fluid coming from a pipe or another source. The gaps in the drawing of field lines exist because we just can't draw all the field lines, passing from every possible point of the paper.
I said volume because it’s area of plate* distance ie field magnitude.
 
  • #7
Delta2
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I said volume because it’s area of plate* distance ie field magnitude.
yes well, I couldn't understand that you see the field magnitude as a distance normal to the surface of the plate.
 
  • #8
rudransh verma
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yes well, I couldn't understand that you see the field magnitude as a distance normal to the surface of the plate.
Here we discuss about flux of uniform field. How to calculate flux of non uniform fields. Is it difficult. There we couldn’t just integrate all the patches. How to calculate that volume of field?
 
  • #9
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Here we discuss about flux of uniform field. How to calculate flux of non uniform fields. Is it difficult. There we couldn’t just integrate all the patches. How to calculate that volume of field?
If the surface is not some simple shape like a plate, and the field is not uniform, then we have to do a surface integral, as post #2 says. I have to say surface integrals with not simple surfaces and not uniform fields, or should I say fields that don,t have some symmetry, are quite hard to compute. What is your mathematical background? Do you know vector calculus?
 
  • #10
rudransh verma
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If the surface is not some simple shape like a plate, and the field is not uniform, then we have to do a surface integral, as post #2 says. I have to say surface integrals with not simple surfaces and not uniform fields, or should I say fields that don,t have some symmetry, are quite hard to compute. What is your mathematical background? Do you know vector calculus?
No
 
  • #11
Delta2
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Well I am afraid you have to learn vector calculus if you want to know how to calculate surface or volume integrals of non uniform vector fields with surfaces or volumes of arbitrary shape. The electric flux is the surface integral of the vector of electric field over a specific surface.
 
  • #12
rudransh verma
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Can you explain:” however we can not do all this by simply comparing the… “ in below image.
 

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  • #13
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Yes, well your book seems to introduce the student to Gauss's law using a "Field line" version of the law, where the charge that is enclosed by a surface is proportional to the number of field lines that pierce the surface. This version of the law is somewhat intuitive or visual I would say.

We need a version of the law that uses mathematical equations to find the enclosed charge ,given the magnitude and direction of the field on the surface, or find the magnitude and direction of the field given the enclosed charge. So just keep reading next section, the electric flux concept is the mathematical way of stating Gauss's law.
 
  • #14
rudransh verma
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Yes, well your book seems to introduce the student to Gauss's law using a "Field line" version of the law, where the charge that is enclosed by a surface is proportional to the number of field lines that pierce the surface. This version of the law is somewhat intuitive or visual I would say.

We need a version of the law that uses mathematical equations to find the enclosed charge ,given the magnitude and direction of the field on the surface, or find the magnitude and direction of the field given the enclosed charge. So just keep reading next section, the electric flux concept is the mathematical way of stating Gauss's law.
Also gauss law is used to simplify the lengthy process of finding the relationship between field and charge for a given Surface. We know the Density of field is proportional to charge. That density is represented by flux and we establish the relationship between density and charge ie gauss law and thus we can find field in easy way of few algebra steps.
 
  • #15
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Just as an aside, when I was a school we learnt the centimetre-gram-second (Gaussian) system of units, where 4pi lines of force originate from a unit charge. Of course, it was emphasised that there are actually no gaps between lines.
 
  • #16
vanhees71
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I think the problem is that the meaning of ##\vec{E} \cdot \mathrm{d} \vec{A}## is not so straight-forward for the electric field.

There's an intuitive meaning for a flow field. In electrodynamics you have electric charge as conserved quantity and in the continuum description (the only natural description within a field theory of course) you have a charge density ##\rho(t,\vec{x})##. This has the operational meaning that when you look at time ##t## at position ##\vec{x}## within a little ("infinitesimal") volume element ##\mathrm{d} V## the amount of charge is ##\mathrm{d} V \rho(t,\vec{x})##. Then this charge of course can move (in fact it's carried around by the charged particles like electrons, protons, and atomic nuclei making up the matter). In the continuuum description this is defined by the current density ##\vec{j}(t,\vec{x})##. The meaning is as follows: Consider at time ##t## a little (infinitesimal) surface with an area ##\mathrm{d} A##. Then define a surface-normal unit vector ##\vec{n}## at the surface (pointing into the one or the other direction; it's up to your choice). Then the amount of charge moving through the surface per unit time is ##\mathrm{d}A \vec{n} \cdot \vec{j}(t,\vec{x})##.

Now consider some volume ##V## with its boundary surface ##\partial V##, which of course is a closed surface. By definition in such a case we define the surface-normal vectors to point out of the volume. Now we can describe charge conservation. The charge contained in ##V## at time ##t## obviously is
$$Q_V(t)=\int_V \mathrm{d} V \rho(t,\vec{x}).$$
Now since total charge is conserved the amount of charge contained in ##V## can only change by some charge flowing into or out of the volume through its surface. The total amount per unit time is
$$\dot{Q}_V(t)=-\int_{\partial V} \mathrm{d} \vec{A} \cdot \vec{j}(t,\vec{x}),$$
and the minus sign comes from the fact that a positive flow integral means that (positive) charge is going out of the volume. That's why this surface is called the flux of the current density.

Of course you can take surface integrals over any kind of vector field and you can use it to express physical laws. One of the fundamental laws of electromagnetism is indeed Gauss's Law for the electric field. It says the "flux" of the electric field through the boundary surface ##\partial V## of the volume ##V## at all times gives the (net) amount of charge contained in the volume
$$\frac{1}{\epsilon_0} Q_V=\int_{\partial V} \mathrm{d} \vec{A} \cdot \vec{E}(t,\vec{x}),$$
and it's called "flux" from the analogy to the case of a current density, where such a surface integral has in a literal sense to do with the flux of the corresponding quantity, but in Gauss's Law there's of course nothing that literally flows somehow.
 
  • #17
rudransh verma
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@Delta2 why do we take flux as negative. It’s not a vector. Is it because this sign is helpful in determining the type of charge in gauss law. -ve flux means -charge and vice versa.
And when the field lines enter the surface from one side and leave the other side, why the flux is taken as 0. It should be a number other than 0 because some amount have pierced the surface. Why do we take flux zero when field lines pass through the surface entirely.
 
  • #18
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@Delta2 why do we take flux as negative. It’s not a vector. Is it because this sign is helpful in determining the type of charge in gauss law. -ve flux means -charge and vice versa.
And when the field lines enter the surface from one side and leave the other side, why the flux is taken as 0. It should be a number other than 0 because some amount have pierced the surface. Why do we take flux zero when field lines pass through the surface entirely.
Yes , flux is not a vector indeed, but it is a scalar, and scalars can be negative or positive.

The sign of the flux has the following interpretation: When our surface is a closed surface then some of the flux will entering the closed area, and this flux is taken as negative, while some other flux might exiting the closed area and this flux that is exiting is taken as positive. If the closed surface doesn't contain any net charge then gauss's law tells us that the total flux is zero, which means that the entering flux (negative flux) is counter balanced by the exiting flux (positive flux).
 
  • #19
vanhees71
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You are right a "flux" or "current" is indeed a scalar quantity. As I tried to explain above it is a quantity describing that something is flowing through a surface. In physics you can take this only literally though if there is really flowing something. The most direct example is that particles move through a surface carrying something with them (e.g., mass or electric charge but also energy, momentum, or angular momentum).

The corresponding vector quantity is a current density, which is in fact a field, ##\vec{j}(t,\vec{x})##. To get the flux or current you first have to specify a surface ##A##. Then you need at any point of the surface a surface-normal vector ##\mathrm{d}^2 \vec{f}##. It's defined by an infinitesimal piece of the surface around the point under consideration and its length is the area of this little surface element. The direction is always perpendicular to the surface, and of course you have two choices, and you can choose any of these two choices you want. This you do for all points (then of course keeping one choice of direction such that the surface element vectors vary smoothly from one point to another). Then the flux or current at time ##t## through the surface is defined as the surface integral
$$\Phi(t)=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{j}(t,\vec{x}).$$
It's indeed a scalar quantity, and its magnitude tells you how much of the quantity under consideration per unit time goes through the surface. The sign of the piece at ##\vec{x}##, ##\mathrm{d} \Phi=\vec{j}(t,\vec{x}) \cdot \mathrm{d}^2 \vec{f}## is determined by the angle between the current density and ##\vec{j}## and ##\vec{j}##. If this angle is between 0 and 90 degrees ##\mathrm{d} \Phi## is positive, i.e., the quantity flows dominantly in the direction chosen when determining the direction of the surface-normal vectors. If the angle is exactly 90 degrees the flow is tangential to the surface at this point, and thus in this case nothing goes through. If the angle is between 90 and 180 degrees it means that the quantity is mainly going in the opposite direction at this point. Integrating over all points of the surface gives the net amount of the quantity flowing through the surface. If it's postive its mainly going in the direction given by the surface-normal vectors. If it's negative it's mainly going in the other direction.

For closed surfaces, which are boundaries of a volume by convention one directs the surface normal elements always pointing out of the volume in question. If you have the flux through such a closed boundary surface a positive sign indicates that the net effect is that the quantity streams out of the volume, if it's negative the net effect is that the quantity streams into the volume.

For Gauss's Law there is not such a direct meaning of something streaming in or out of the surface. The vector field integrated over the closed boundary surface of the volume here is the electric field, which doesn't describe some quantity flowing around. It's just defined by the force acting on a point charge located at the point under consideration ##\vec{F}=q \vec{E}(t,\vec{x})##, where ##q## is the electric charge of the test particle. Gauss's Law then states that the flux of the electric field, i.e., the corresponding surface integral, is (modulo a factor of the used system of units, in the SI it's ##1/\epsilon_0## needed for unit conversion) is the charge contained in the volume (at the time ##t##)
$$Q_V(t)=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}(t,\vec{x}).$$
This is one of the fundamental laws of electromagnetism (i.e., one of Maxwell's equations) and it cannot be derived. It's justified only, because together with the other Maxwell equations it describes all phenomena related with the electromagnetic field, manifesting itself through the forces acting on charged bodies/particles.
 
  • #20
rudransh verma
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Yes , flux is not a vector indeed, but it is a scalar, and scalars can be negative or positive.

The sign of the flux has the following interpretation: When our surface is a closed surface then some of the flux will entering the closed area, and this flux is taken as negative, while some other flux might exiting the closed area and this flux that is exiting is taken as positive. If the closed surface doesn't contain any net charge then gauss's law tells us that the total flux is zero, which means that the entering flux (negative flux) is counter balanced by the exiting flux (positive flux).
So you are saying flux is like a force that gets balanced. Doesn’t make sense. We know flux is the amount of lines piercing. So if a surface is placed in a field then it will get pierced by it. So there must be some flux even if there is no net charge.
 
  • #21
Delta2
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So you are saying flux is like a force that gets balanced. Doesn’t make sense. We know flux is the amount of lines piercing. So if a surface is placed in a field then it will get pierced by it. So there must be some flux even if there is no net charge.
You take the piercing too literally. Even so just think then that there is negative piercing and positive piercing...
 
  • #22
rudransh verma
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You take the piercing too literally. Even so just think then that there is negative piercing and positive piercing...
Book says because the field lines of field all pass from one end to another of a cylinder. So the net flux here is zero. Mathematically one end has +flux and other has -flux. Net flux is zero. So it means for the net flux to be nonzero there should be more lines entering the surface than leaving or vice versa. It’s not like if the field pierces then there should be a flux. It’s also necessary that they should terminate or originate from inside the surface. Not pass from inside the surface.Right?
 
  • #23
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Book says because the field lines of field all pass from one end to another of a cylinder. So the net flux here is zero. Mathematically one end has +flux and other has -flux. Net flux is zero. So it means for the net flux to be nonzero there should be more lines entering the surface than leaving or vice versa. It’s not like if the field pierces then there should be a flux
That looks good, seems to me you got it right!
It’s also necessary that they should terminate or originate from inside the surface.
If the whole space (whole universe) contains charges only inside our closed surface then this is correct.
Not pass from inside the surface.Right?
But the whole space might contain charges only outside our closed surface, which means in this case that the field lines just pass through our closed surface, the flux in this case will be positive in some areas of our closed surface (in the areas where the field lines exit), and negative in some other areas (where the field lines enter). The total flux will be zero in this case.
 
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  • #24
rudransh verma
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If the whole space (whole universe) contains charges only inside our closed surface then this is correct.
I didn’t get it! Please elaborate.
 
  • #25
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I didn’t get it! Please elaborate.
What exactly you didnt get? What happens if all the charges are inside our closed surface?
 

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