Question regarding the use of Electric flux and Field Lines

In summary, the concept of electric field lines is often misunderstood, as they are not the same as electric flux. Electric flux is defined as the integral of the electric field over a closed surface, and it is proportional to the total charge enclosed by that surface. The use of multiplying the electric field by the area may seem arbitrary, but it is a restatement of one of Maxwell's equations and has practical applications in calculations. In cgs units, the constant 1/4πε0 is set to 1, causing the electric flux to be equal to 4π times the enclosed charge.
  • #1
Harikesh_33
24
4
1)Field Lines is supposed to represent the electric field around a charge ,now we can draw infinite field lines around a charge and sinc Electric flux is No of Field Lines /area ,does it become infinite ,the whole concept of field lines is quite in the Gray Area for me ,I can in theory mark several points around a charge and draw the electric field vector in all those points ,what's stopping me from joining all those points and drawing electric field lines that passes through all of them .

2) I recently read in a Quora post that the Electrostatic force in Gaussian CGS units is just ##\dfrac{q_1q_2}{r^2}## ,whole converting to SI units we get a factor of ##ε_0## for free space ,now does that mean in CGS units the flux is just around a charge is just Q( here the net charge is Q) because in SI units our formula becomes ##\dfrac{Q}{ε_0}## ,What I am meaning to say is I don't the whole concept of multiplying E field vecotr and infinitesimal area to get the infinitesimal flux for that infinitesimal area ,multiplying by dA reduces our E field value ,does this mean our E field vector is per unit area (if dA =0.0000001,then when multiplied with E.##\vec{n}## this term becomes small ) does this mean our E field vector is defined per unit area

.https://www.quora.com/What-is-an-intuitive-explanation-of-the-Vacuum-permittivity/answer/Thomas-Jollans?ch=10&oid=5917841&share=6d7771dd&srid=pGwHF&target_type=answer
 
Physics news on Phys.org
  • #2
Harikesh_33 said:
sinc Electric flux is No of Field Lines /area
It's not - it's ##\int_A \vec E\cdot d\vec A##. Under some assumptions about how many field lines and which ones you drew this is proportional to the number of lines piercing a surface, yes, but that is not the definition of flux.
Harikesh_33 said:
multiplying by dA reduces our E field value
No - dA is infinitesimal.

What's going on here is that we split our surface up into a lot of small polygons of area ##\delta A##, calculate ##E\delta A## for each one, and sum all those terms to get a total flux. This gets more exact the smaller the polygons, and polygons of area ##dA## are the limit as they tend to infinitely small. We call that an integral rather than a sum, but it's the same principle.
 
  • Like
Likes vanhees71
  • #3
Ibix said:
It's not - it's ##\int_A \vec E\cdot d\vec A##. Under some assumptions about how many field lines and which ones you drew this is proportional to the number of lines piercing a surface, yes, but that is not the definition of flux.

No - dA is infinitesimal.

What's going on here is that we split our surface up into a lot of small polygons of area ##\delta A##, calculate ##E\delta A## for each one, and sum all those terms to get a total flux. This gets more exact the smaller the polygons, and polygons of area ##dA## are the limit as they tend to infinitely small. We call that an integral rather than a sum, but it's the same principle.
I get the math part ,my question was why multiply area and electric field ?Why is flux defined as Electric field *Area,what is the meaning of Area * electric field ,What other quantity has its unit as Nm²/C?
 
  • #4
Because it's useful to do so. Integrating over a closed surface gives you the integral form of one of Maxwell's equations, for example.
 
  • Like
Likes robphy and vanhees71
  • #5
When we learned cgs units we were told that Gauss said that 4 pi lines of force originate from unit charge. The cgs unit of flux being called the Line.
 
  • #6
Ibix said:
Because it's useful to do so. Integrating over a closed surface gives you the integral form of one of Maxwell's equations, for example.
I get that it's useful to do ,but isn't there a meaning for that ?
 
  • #7
tech99 said:
When we learned cgs units we were told that Gauss said that 4 pi lines of force originate from unit charge. The cgs unit of flux being called the Line.
Could you elaborate as to how 4π lines originate ,does it have to do something with Surface area of Sphere ?
 
  • #8
Gauss' Law says "the electric flux through any closed surface is proportional to the total charge enclosed by that surface".

Possibly enlightening:


I think of an "electric field line" (in a diagram using the "density of electric field lines" to represent the relative strength of the electric field) as a representation of a "unit of flux".
 
  • #9
Harikesh_33 said:
I get that it's useful to do ,but isn't there a meaning for that ?
Well, it's a restatement of a Maxwell equation, so the meaning is the same as ##\vec\nabla\cdot\vec E=\rho_e##, that field lines can only originate on charges.
Harikesh_33 said:
Could you elaborate as to how 4π lines originate ,does it have to do something with Surface area of Sphere ?
No - it's just that in cgs units you set ##1/4\pi\epsilon_0=1## so that ##F=qQ/r^2## as you noted in your OP. That makes the SI ##Q/\epsilon_0## become ##4\pi Q## in cgs (not ##Q## as you stated).
 
Last edited:
  • Like
Likes tech99
  • #10
Harikesh_33 said:
I get that it's useful to do ,but isn't there a meaning for that ?
It's one of the fundamental Maxwell equations. In integral form it reads (in SI units)
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} Q_V,$$
where ##V## is an arbitrary volume and ##\partial V## its boundary surface. It says that the electric field's flux through a closed boundary equals the charge contained in that volume (multiplied by ##1/\epsilon_0##, which is a conversion factor of the used SI units, which define the unit of electric charge, the Coulomb (C), by setting the elementary charge to a certain value of about ##1.6 \cdot 10^{-19} \text{C}##).

This is a fundamental law of Nature, which you cannot derive from simpler fundamental principles. Historically it has been empirically found.

Also it's completely independent of the choice of the unit system. The SI is the officially used system of units in (experimental) physics. Gaussian or rationalized Gaussian (Heaviside-Lorentz) units are still in use in theoretical physics due to their more "natural" character concerning the dimensions of the electromagnetic quantities (e.g., ##\vec{E}## and ##\vec{B}## have the same units, which is "natural", because from a relativistic point of view these are just the components of the one and only electromagnetic field, described as an antisymmetric 2nd-rank Minkowski tensor).
 
  • #11
vanhees71 said:
It's one of the fundamental Maxwell equations. In integral form it reads (in SI units)
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} Q_V,$$
where ##V## is an arbitrary volume and ##\partial V## its boundary surface. It says that the electric field's flux through a closed boundary equals the charge contained in that volume (multiplied by ##1/\epsilon_0##, which is a conversion factor of the used SI units, which define the unit of electric charge, the Coulomb (C), by setting the elementary charge to a certain value of about ##1.6 \cdot 10^{-19} \text{C}##).

This is a fundamental law of Nature, which you cannot derive from simpler fundamental principles. Historically it has been empirically found.

Also it's completely independent of the choice of the unit system. The SI is the officially used system of units in (experimental) physics. Gaussian or rationalized Gaussian (Heaviside-Lorentz) units are still in use in theoretical physics due to their more "natural" character concerning the dimensions of the electromagnetic quantities (e.g., ##\vec{E}## and ##\vec{B}## have the same units, which is "natural", because from a relativistic point of view these are just the components of the one and only electromagnetic field, described as an antisymmetric 2nd-rank Minkowski tensor).
My problems are these
1)
A charge experiences force when placed on the path of field lines ,Iknow that the electric field is large where the density of field lines are large ,then I know that the Electric field is tangential to field line and a charge experiences a force of magnitude qE in that direction ,now does this mean that electric field lines carry force equal to that value of Electric field vector at that point ,(or) is it due to the quantity of field lines at a point .

2)Say the dot product of E field and normal vector is 5 ,to find the E field passing through the area dA (0.0001) ,we write it as E.ndA ,now 5 * 0.0001 =0.0005 ,we had the E field at a point the area of a point is less than 0.0001 m² ,now multiply by anything that has its area more than the point must increase the E field in that area ,what is actually happening here ?
 
  • #12
Harikesh_33 said:
My problems are these
1)
A charge experiences force when placed on the path of field lines ,Iknow that the electric field is large where the density of field lines are large ,then I know that the Electric field is tangential to field line and a charge experiences a force of magnitude qE in that direction ,now does this mean that electric field lines carry force equal to that value of Electric field vector at that point ,(or) is it due to the quantity of field lines at a point .

2)Say the dot product of E field and normal vector is 5 ,to find the E field passing through the area dA (0.0001) ,we write it as E.ndA ,now 5 * 0.0001 =0.0005 ,we had the E field at a point the area of a point is less than 0.0001 m² ,now multiply by anything that has its area more than the point must increase the E field in that area ,what is actually happening here ?
So Gauss's Law in Gaussian units is just ##\int_s{\vec{E}.dA}=Q##?
 
  • #13
Harikesh_33 said:
My problems are these
1)
A charge experiences force when placed on the path of field lines ,Iknow that the electric field is large where the density of field lines are large ,then I know that the Electric field is tangential to field line and a charge experiences a force of magnitude qE in that direction ,now does this mean that electric field lines carry force equal to that value of Electric field vector at that point ,(or) is it due to the quantity of field lines at a point .

2)Say the dot product of E field and normal vector is 5 ,to find the E field passing through the area dA (0.0001) ,we write it as E.ndA ,now 5 * 0.0001 =0.0005 ,we had the E field at a point the area of a point is less than 0.0001 m² ,now multiply by anything that has its area more than the point must increase the E field in that area ,what is actually happening here ?
Forget about "field lines" and think in terms of the "field" itself. The em. field is around any electrically charged particle (electric charge being a fundamental property of matter). The electric force on another charged particle is due to the presence of the electromagnetic field at the position of this particle.

This field-poin-of-view, introduced by Faraday in the early 19th century, was a very important step towards the modern understanding of the fundamental interactions from the "action-at-a-distance paradigm" of Newton to a local description.

Today it's clear that the electromagnetic field is just one of several fundamental fields describing (in it's quantum-theoretical formulation) both the interactions between particles and the particles themselves.

I don't understand what you mean by item 2).
 
  • #14
Harikesh_33 said:
So Gauss's Law in Gaussian units is just ##\int_s{\vec{E}.dA}=Q##?
In Gaussian units its ##4 \pi Q## on the right-hand side.
 
  • Like
Likes tech99
  • #15
vanhees71 said:
Forget about "field lines" and think in terms of the "field" itself. The em. field is around any electrically charged particle (electric charge being a fundamental property of matter). The electric force on another charged particle is due to the presence of the electromagnetic field at the position of this particle.

This field-poin-of-view, introduced by Faraday in the early 19th century, was a very important step towards the modern understanding of the fundamental interactions from the "action-at-a-distance paradigm" of Newton to a local description.

Today it's clear that the electromagnetic field is just one of several fundamental fields describing (in it's quantum-theoretical formulation) both the interactions between particles and the particles themselves.

I don't understand what you mean by item 2).
I meant the following :
Say you are tasked with finding the mass of a body ,with the density and volume given ,consider Lead for example (Lead has its density as 11,300 kg/m³ at a point A,this means that if you repeat the mass distribution of A until the point becomes a volume of 1m³ then your mass will be 11,300 kg .

Now we have this out of the way ,say your volume is 0.01m³ ,now to find the massof that volume ,we multiply density and volume ,(ie) we get 11,300kg/m³*0.01m³ =113 kg (if you see here density > mass ) .

Now similarly if we are tasked with finding ##\vec{E}.\hat{n}dA## ,and the quantity ##E.\hat{n}## is the electric field along the surface dA at a point ,say that ##\vec{E}.\hat{n}## is 5 at a point ,our dA is ##0.01m²## now ##\vec{E}\hat{n}.dA## is 0.05 ,now as we see here multiplying the magnitude of the area reduces the field value like the density example ,Now does this mean that our E field at a point is defined for an area ?
 
  • #16
The electric field is a vector at any point in space at a given time, ##\vec{E}(t,\vec{x})##. I still don't understand, where your problem is.
 
  • #17
vanhees71 said:
The electric field is a vector at any point in space at a given time, ##\vec{E}(t,\vec{x})##. I still don't understand, where your problem is.
I am talking about the Time Independent One .

Ok ,my problem is ,E field is defined for a point {a object with no area },say at some point A it's 5 ,(let's assume that the E field vector and normal vector are along the same direction here ),now when we multiply a area that is greater than the point A 's area (say 0.01m²) ,shouldn't the value of the flux at that point be greater the field at that point ?
 
  • #18
The electric field is defined at each point in space. I still don't understand the question. The flux has dimension "electric field times area", i.e., Vm. How do you think you can compare it with the (magnitude of the) electric field which has dimension V/m? That simply doesn't make sense.
 
  • Like
Likes robphy
  • #19
vanhees71 said:
The electric field is defined at each point in space. I still don't understand the question. The flux has dimension "electric field times area", i.e., Vm. How do you think you can compare it with the (magnitude of the) electric field which has dimension V/m? That simply doesn't make sense.
Ok what does the flux at a point tell us about the E field at a point ?
 
  • #20
Harikesh_33 said:
Ok what does the flux at a point tell us about the E field at a point ?
Flux is defined through a surface, not at a point. And it may well tell you very little about the E field. However, knowing the flux through some collection of surfaces will let you deduce the E field in the same way knowing the divergence of the E field everywhere (plus a boundary condition) let's you deduce the E field.
 
  • Like
Likes vanhees71 and Harikesh_33
  • #21
Ibix said:
Flux is defined through a surface, not at a point. And it may well tell you very little about the E field. However, knowing the flux through some collection of surfaces will let you deduce the E field in the same way knowing the divergence of the E field everywhere (plus a boundary condition) let's you deduce the E field.
So flux itself doesn't have a very good meaning ? It's only purpose is to find Electric Field ,but why is the flux equal to 4πQ intuitively(Gaussian units) lime you mentioned before ?
 
  • #22
Not too much. You need a closed surface around a point. You can make the enclosed volume smaller and smaller. Dividing by this volume contracting it to a point in this sense, leads to the really fundamental form of the corresponding Maxwell equation (Gauss's Law for the electric field),
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
 
  • #23
Harikesh_33 said:
why is the flux equal to 4πQ intuitively(Gaussian units) lime you mentioned before ?
It's one of Maxwell's equations. I'm not sure there's an answer to "why" here, beyond observing that this is the classical limit of the behaviour of a massless spin-1 quantum field. Why there's a massless spin-1 quantum field in the first place is, of course, an open question.
 
  • Like
Likes vanhees71 and ergospherical
  • #24
You can obtain Gauss' law by setting the variation of the action ##S = S_{\mathrm{field}} + S_{\mathrm{interaction}} = -\int (F^2/(16 \pi c) + A_{\mu} j^{\mu} /c^2) d^4 x## to zero. (Here only the potentials are varied, whilst the particle trajectory is assumed given).

See: Landau & Lifshitz, Classical Theory of Fields.
 
  • Like
Likes vanhees71 and Ibix
  • #25
Harikesh_33 said:
So flux itself doesn't have a very good meaning ? It's only purpose is to find Electric Field ,but why is the flux equal to 4πQ intuitively(Gaussian units) lime you mentioned before ?
The Flux of a Vector Field is ##\iint \vec V \cdot d\vec A## is akin to a line integral ##\int \vec V \cdot d\vec \ell##
in the sense that it is a measure of the component of the vector field:
perpendicular to a surface element (for the flux-integral)
and parallel to a section of a curve (for the line-integral).
These constructions from mathematics show up in the description of some physical laws.

The flux of a vector field shows up in all 4 of Maxwell's equations and in the continuity equation:

Gauss for E: ##{\LARGE\unicode{x222F}}_{\partial V} \vec E \cdot d\vec A =\frac{1}{\epsilon_0}{\Large\iiint}_V\ \rho\ dV##
Gauss for B: ##{\LARGE\unicode{x222F}}_{\partial V} \vec B \cdot d\vec A =0##
Ampere-Maxwell: ##{\Large\oint}_{\partial A} \vec B \cdot d\vec \ell
=\mu_0 {\Large\iint}_A\ \vec \jmath\cdot d\vec A +\mu_0\left(\epsilon_0 \frac{d}{dt}{\Large\iint}_A \vec E \cdot d\vec A\right)##
Faraday: ##{\Large\oint}_{\partial A} \vec E \cdot d\vec \ell
=- \frac{d}{dt}{\Large\iint}_A \vec B \cdot d\vec A##
[... and there was light!]continuity: ##\frac{d}{dt} {\Large\iiint}_V\ \rho\ dV + {\LARGE\unicode{x222F}}_{\partial V}\ \vec \jmath \cdot d\vec A=0##

(I tell my students:
the Gauss Law for E is always true... but not always useful for finding the electric field in general situations.)Possibly useful... although it makes reference to differential forms (but maybe that's not a bad thing).
Listen to the "ideas".





(start at 12m00s)

(start at 12m00s)
 
Last edited:
  • Like
Likes vanhees71 and Ibix
  • #26
robphy said:
(I tell my students:
the Gauss Law for E is always true... but not always useful for finding the electric field in general situations.)
Well, the Maxwell equations in integral form are almost never applicable to find solutions of these equations. Only for very symmetric cases you may use them. Anyway, the true laws are the local differential Maxwell equations, and then I stress the locality as the fundamental reason for introducing the "field paradigm" (historically by Faraday and then of course Maxwell) and its necessity in connection with the fact that the world is described by relativistic spacetime models rather than the only approximately valid Newtonian spacetime model, where "actions at a distance" are consistent, which is no longer the case when it comes to relativistic situations, and E&M is almost always relativistic (although amazingly you can get a long way in electrodynamics without even noticing it, but paying the prize of getting a pretty enigmatic picture of the phenomena ;-)).
 
  • Like
Likes Ibix

Related to Question regarding the use of Electric flux and Field Lines

1. What is electric flux and how is it related to electric field lines?

Electric flux is a measure of the amount of electric field passing through a given area. It is directly related to the number of electric field lines passing through that area. The more field lines passing through an area, the higher the electric flux will be.

2. How is electric flux calculated?

Electric flux is calculated by taking the dot product of the electric field and the area vector. This means multiplying the magnitude of the electric field by the magnitude of the area and then multiplying that by the cosine of the angle between the two vectors.

3. What are electric field lines and how are they represented?

Electric field lines are imaginary lines that represent the direction and strength of an electric field. They are drawn from positive to negative charges and the density of the lines represents the strength of the field. The closer the lines are together, the stronger the field is.

4. Can electric field lines cross each other?

No, electric field lines cannot cross each other. This is because the direction of the electric field at any point is determined by the tangent to the field line at that point. If the lines were to cross, the direction of the field would be ambiguous.

5. How can electric field lines be used to visualize electric fields?

Electric field lines can be used to visualize electric fields by showing the direction and strength of the field at different points. The lines can also be used to determine the net electric field at a point by looking at the direction and magnitude of the lines in that area.

Similar threads

Replies
35
Views
3K
  • Electromagnetism
Replies
30
Views
2K
  • Electromagnetism
Replies
2
Views
4K
  • Electromagnetism
Replies
2
Views
2K
Replies
11
Views
2K
Replies
3
Views
2K
  • Electromagnetism
2
Replies
51
Views
6K
  • Electromagnetism
Replies
1
Views
1K
Replies
2
Views
1K
Replies
8
Views
978
Back
Top