Maxwell's Two Equations For Electrostatic Fields

This physics video tutorial provides a basic introduction into maxwell's equations and electromagnetic waves. Maxwell's 4 equations include gauss' law for el. Electrostatic Fields Electrostatic fields are static (time-invariant) electric fields produced by static (stationary) charges. The mathematical definition of the electrostatic field is derived from Coulomb’s law which defines the vector force between two point charges. Coulomb’s Law Given point charges q 1, q 2 (units=C). The displacement current introduced by Maxwell results instead from a changing electric field and accounts for a changing electric field producing a magnetic field. The equations for the effects of both changing electric fields and changing magnetic fields differ in form only where the absence of magnetic monopoles leads to missing terms. In Maxwell's equations. The behavior of magnetic fields (B, H), electric fields (E, D), charge density (ρ), and current density (J) is described by Maxwell's equations. The role of the magnetization is described below. Relations between B, H, and M. In Gauss's law, the electric field is the electrostatic field. The law shows how the electrostatic field behaves and varies depending on the charge distribution within it. More formally it relates the electric flux the electric field flowing from positive to negative charges passing through a closed surface to the charge contained within the.

Michael Fowler, PhysicsDepartment, UVa

The Equations

Maxwell’s four equations describe the electric and magneticfields arising from distributions of electric charges and currents, and howthose fields change in time. They were the mathematical distillation ofdecades of experimental observations of the electric and magnetic effects ofcharges and currents, plus the profound intuition of Michael Faraday. Maxwell’s own contribution to these equationsis just the last term of the last equation—but theaddition of that term had dramatic consequences. It made evident for the first time thatvarying electric and magnetic fields could feed off each other—thesefields could propagate indefinitely through space, far from the varying chargesand currents where they originated. Previously these fields had been envisioned as tethered to the chargesand currents giving rise to them. Maxwell’snew term (called the displacement current) freed them to move through space ina self-sustaining fashion, and even predicted their velocity—it was thevelocity of light!

Here are the equations:

1. Gauss’ Law for electric fields:

Maxwell's Two Equations For Electrostatic Fields Based

$\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}=q/{\epsilon }_{0}.$

(The integral of the outgoing electric field over an area enclosing a volume equals the total charge inside, in appropriate units.)

2. The corresponding formula for magnetic fields:

$\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}=0.$

(No magnetic charge exists: no “monopoles”.)

3. Faraday’s Law of Magnetic Induction:

$\oint \stackrel{\to }{E}\cdot d\stackrel{\to }{\ell }=-d/dt\left(\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}\right).$

The first term is integrated round a closed line, usually a wire, and gives the total voltage change around the circuit, which is generated by a varying magnetic field threading through the circuit.

4. Ampere’s Law plus Maxwell’s displacement current:

$\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}\left(I+\frac{d}{dt}\left({\epsilon }_{0}\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}\right)\right).$

This gives the total magnetic force around a circuit in terms of the current through the circuit, plus any varying electric field through the circuit (that’s the “displacement current”).

The purpose of this lecture is toreview the first three equations and the original Ampere’s law fairly briefly,as they were covered earlier in the course, then to demonstrate why thedisplacement current term must be added for consistency, and finally to show,without using differential equations, how measured values of static electricaland magnetic attraction are sufficient to determine the speed of light.

Preliminaries: Definitions of µ0and ε0

Ampere discovered that two parallel wires carrying electriccurrents in the same direction attract each other magnetically, the force innewtons per unit length being given by

$F=2\left(\frac{{\mu }_{0}}{4\pi }\right)\frac{{I}_{1}{I}_{2}}{r},$

for long wires a distance $r$ apart. We are using the standard modern units(SI). The constant ${\mu }_{0}/4\pi$ that appears here is exactly 10-7,this defines our present unit of current, the ampere. To repeat: ${\mu }_{0}/4\pi$ is not something to measure experimentally,it's just a funny way of writing the number 10-7! That's not quite fair—it hasdimensions to ensure that both sides of the above equation have the samedimensionality. (Of course, there's a historical reason for this strangeconvention, as we shall see later). Anyway,if we bear in mind that dimensions have been taken care of, and justwrite the equation

$F=2\cdot {10}^{-7}\cdot \frac{{I}_{1}{I}_{2}}{r},$

it's clear that this defines the unit current—one ampere—as thatcurrent in a long straight wire which exerts a magnetic force of $2×{10}^{-7}$ newtons per meter of wire on a parallel wireone meter away carrying the same current.

However, after we have established our unit of current—the amperewehave also thereby defined our unit of charge, since current is a flow ofcharge, and the unit of charge must be the amount carried past a fixed point inunit time by unit current. Therefore,our unit of charge—the coulomb—is definedby stating that a one amp current in a wire carries one coulombper second past a fixed point.

To be consistent, we must do electrostatics using thissame unit of charge. Now, the electrostatic force between two chargesis $\left(1/4\pi {\epsilon }_{0}\right){q}_{1}{q}_{2}/{r}^{2}.$ The constant appearing here, now written $1/4\pi {\epsilon }_{0}$,must be experimentally measured—its valueturns out to be $9×{10}^{9}$.

To summarize: tofind the value of $1/4\pi {\epsilon }_{0}$,two experiments have to be performed.We must first establish the unit of charge from the unit of current bymeasuring the magnetic force between two current-carrying parallel wires. Second, we must find the electrostatic forcebetween measured charges. (We could, alternatively, have defined some otherunit of current from the start, then we would have had to find both ${\mu }_{0}$ and ${\epsilon }_{0}$ by experiments on magnetic and electrostaticattraction. In fact, the ampere was originallydefined as the current that deposited a definite weight of silver per hour inan electrolytic cell).

Maxwell's Equations

We have so far established that the total flux of electricfield out of a closed surface is just the total enclosed charge multiplied by $1/{\epsilon }_{0}$,

$\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}=q/{\epsilon }_{0}.$

This is Maxwell’sfirst equation. Itrepresents completely covering the surface with a large number of tiny patcheshaving areas $d\stackrel{\to }{A}$. We represent these small areas asvectors pointing outwards, because we can then take the dot product with theelectric field to select the component of that field pointing perpendicularlyoutwards (it would count negatively if the field were pointing inwards)—this isthe only component of the field that contributes to actual flow across thesurface. (Just as a river flowingparallel to its banks has no flow across the banks).

The second Maxwell equation is the analogous one forthe magnetic field, which has no sources or sinks (no magnetic monopoles, thefield lines just flow around in closed curves). Thinking of the force lines asrepresenting a kind of fluid flow, the so-called 'magnetic flux', wesee that for a closed surface, as much magnetic flux flows into the surface asflows out. This can perhaps bevisualized most clearly by taking a group of neighboring lines of force forminga slender tube—the'fluid' inside this tube flows round and round, so as the tube goesinto the closed surface then comes out again (maybe more than once) it is easyto see that what flows into the closed surface at one place flows out at another.Therefore the net flux out of the enclosed volume is zero, Maxwell’s second equation:

$\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}=0.$

The first two Maxwell's equations, given above, are forintegrals of the electric and magnetic fields over closed surfaces .The other two Maxwell's equations, discussed below, are for integrals ofelectric and magnetic fields around closed curves (taking thecomponent of the field pointing along the curve). These represent the work thatwould be needed to take a charge around a closed curve in an electric field,and a magnetic monopole (if one existed!) around a closed curve in a magneticfield.

The simplest version of Maxwell's third equationis the electrostatic case:

The path integral$\oint \stackrel{\to }{E}\cdot d\stackrel{\to }{\ell }=0$for electrostatics.

However, we know that this is only part of the truth,because from Faraday's Law of Induction, if a closed circuit has a changingmagnetic flux through it, a circulating current will arise, which meansthere is a nonzero voltage around the circuit.

The full version of Maxwell's third equation is:

$\oint \stackrel{\to }{E}\cdot d\stackrel{\to }{\ell }=-d/dt\left(\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}\right)$

where the area integrated over on the right hand side spansthe path (or circuit) on the left hand side, like a soap film on a loop ofwire.

It may seem that the integral on the right hand sideis not very clearly defined, because if the path or circuit lies in a plane,the natural choice of spanning surface (the 'soap film') is flat, buthow do you decide what surface to choose to do the integral over for a wirebent into a circuit that doesn’t lie in a plane? The answer is that it doesn’t matter whatsurface you choose, as long as the wire forms its boundary. Consider two different surfaces both havingthe wire as a boundary (just as both the northern hemisphere of the earth’ssurface and the southern hemisphere have the equator as a boundary). If you addthese two surfaces together, they form a single closed surface, and we knowthat for a closed surface $\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}=0$. This implies that $\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}$ for one of the two surfaces bounded by thepath is equal to $-\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}$ for the other one, so that the two will add tozero for the whole closed surface. Butdon’t forget these integrals for the whole closed surface are defined with thelittle area vectors pointing outwards from the enclosed volume. By imaginingtwo surfaces spanning the wire that are actually close to each other, it isclear that the integral over one of them is equal to the integral over theother if we take the $d\stackrel{\to }{A}$ vectors to point in the same direction forboth of them, which in terms of the enclosed volume would be outwards for onesurface, inwards for the other one. The bottom line of all this is that thesurface integral $\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}$ is the same for any surface spanning the path,so it doesn’t matter which we choose.

The equation analogous to the electrostatic version of thethird equation given above, but for the magnetic field, is Ampere's law

$\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}\cdot \text{(enclosed currents)}$

for magnetostatics, where the currents counted are those threading throughthe path we're integrating around, so if there is a soap film spanning thepath, these are the currents that punch through the film (of course, we have toagree on a direction, and subtract currents flowing in the opposite direction).

We must now consider whether this equation, like theelectrostatic one, has limited validity. In fact, it was not questioned for ageneration after Ampere wrote it down: Maxwell's great contribution, in the1860's, was to realize that it was not always valid.

When Does Ampere'sLaw Go Wrong?

A simple example to see that something mustbe wrong with Ampere's Law in the general case is given by Feynman in his Lecturesin Physics. Suppose we use ahypodermic needle to insert a spherically symmetric blob of charge in themiddle of a large vat of solidified jello (which we assume conductselectricity). Because of electrostaticrepulsion, the charge will dissipate, currents will flow outwards in aspherically symmetric way. Question: does this outward-flowingcurrent distribution generate a magnetic field? The answer must be no , because sincewe have a completely spherically symmetric situation, it could only generate aspherically symmetric magnetic field. Butthe only possible such fields are one pointing outwards everywhere and onepointing inwards everywhere, both corresponding to non-existent monopoles. So,there can be no magnetic field.

However,imagine we now consider checking Ampere's law by taking as a path a horizontalcircle with its center above the point where we injected the charge (think of ahalo above someone’s head.) Obviously,the left hand side of Ampere's equation is zero, since there can be no magneticfield. (It would have to be sphericallysymmetric, meaning radial.) On the otherhand, the right hand side is most definitely not zero, since some of theoutward flowing current is going to go through our circle. So the equation must be wrong.

Ampere's law was established as the result of largenumbers of careful experiments on all kinds of current distributions. So how could it be that something of the kindwe describe above was overlooked? Thereason is really similar to why electromagnetic induction was missed for solong. No-one thought about looking at changing fields, all theexperiments were done on steady situations. With our ball of charge spreading outward inthe jello, there is obviously a changing electric field. Imagine yourself in the jello near where thecharge was injected: at first, you would feel a strong field from the nearbyconcentrated charge, but as the charge spreads out spherically, some of itgoing past you, the field will decrease with time.

Maxwell's Example

Maxwell himself gave a more practical example: consider Ampere's law for the usual infinitelylong wire carrying a steady current $I,$ but now break the wire at some point and putin two large circular metal plates, a capacitor, maintaining the steady current$I$ in the wire everywhere else, so that charge issimply piling up on one of the plates and draining off the other.

Looking now at the wire some distance away from theplates, the situation appears normal, and if we put the usual circular patharound the wire, application of Ampere's law tells us that the magnetic fieldat distance $r,$ from

$\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}I$

is just

$B={\mu }_{0}I/2\pi r.$

Recall, however, that wedefined the current threading the path in terms of current punching through asoap film spanning the path, and said this was independent of whether the soapfilm was flat, bulging out on one side, or whatever. With a single infinitewire, there was no escape— no contortions of this coveringsurface could wriggle free of the wire going through it (actually, if youdistort the surface enough, the wire could penetrate it several times, but youhave to count the net flow across the surface, and the new penetrations wouldcome in pairs with the current crossing the surface in opposite directions, sothey would cancel).

Oncewe bring in Maxwell's parallel plate capacitor, however, there is away to distort the surface so that no current penetrates it at all: we can runit between the plates!  The question then arises: can we rescue Ampere's law byadding another term just as the electrostatic version of the third equation wasrescued by adding Faraday's induction term? The answer is of course yes:although there is no current crossing the surface if we put it betweenthe capacitor plates, there is certainly a changing electric field ,because the capacitor is charging up as the current $I$ flows in. Assuming the plates are closetogether, we can take all the electric field lines from the charge $q$ on one plate to flow across to the otherplate, so the total electric flux across the surface between the plates,

$\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}=q/{\epsilon }_{0}.$

Now, the current in the wire $,I,$ is just the rate of change of charge on theplate,

$I=dq/dt.$

Putting the above two equations together, we see that

$I=\frac{d}{dt}\left({\epsilon }_{0}\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}\right).$

Ampere's law can now be written in a way that is correct nomatter where we put the surface spanning the path we integrate the magneticfield around:

$\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}\left(I+\frac{d}{dt}\left({\epsilon }_{0}\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}\right)\right).$

This is Maxwell’sfourth equation.

Notice that in the case of the wire, either thecurrent in the wire, or the increasing electric field, contribute onthe right hand side, depending on whether we have the surface simply cuttingthrough the wire, or positioned between the plates. (Actually, more complicatedsituations are possible—we couldimaging the surface partly between the plates, then cutting through the platesto get out! In this case, we would have tofigure out the current actually in the plate to get the right hand side, butthe equation would still apply).

'DisplacementCurrent'

Maxwell referred to the second term on the right handside, the changing electric field term, as the 'displacement current'. This was an analogy with a dielectricmaterial. If a dielectric material isplaced in an electric field, the molecules are distorted, their positivecharges moving slightly to the right, say, the negative charges slightly to theleft. Now consider what happens to adielectric in an increasing electric field. The positive charges will be displaced to theright by a continuously increasing distance, so, as long as the electric fieldis increasing in strength, these charges are moving: there is actually a displacementcurrent . (Meanwhile, the negativecharges are moving the other way, but that is a current in the same direction,so adds to the effect of the positive charges' motion.) Maxwell's picture of the vacuum, the aether,was that it too had dielectric properties somehow, so he pictured a similarmotion of charge in the vacuum to that we have just described in thedielectric. The picture is wrong, but thisis why the changing electric field term is often called the 'displacementcurrent', and in Ampere's law (generalized) is just added to the realcurrent, to give Maxwell's fourth—and final—equation.

Another Angle on theFourth Equation: the Link to Charge Conservation

Going back for a moment to Ampere's law, we stated itas:

$\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}\cdot \text{(enclosed currents)}$

for magnetostatics, where the currents counted are those threading through thepath we’re integrating around, so if there is a soap film spanning the path,these are the currents that punch through the film. Our mental picture here isusually of a few thin wires, maybe twisted in various ways, carrying currents.More generally, thinking of electrolytes, or even of fat wires, we should beenvisioning a current density varying from point to point in space. In otherwords, we have a flux of current and the natural expression for the currentthreading our path is (analogous to the magnetic flux in the third equation) towrite a surface integral of the current density $\stackrel{\to }{j}$ over a surface spanning the path, giving formagnetostatics

path integral $\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}$(surface integral, over surface spanning path)

The question then arises as to whether the surface integralwe have written on the right hand side above depends on which surfacewe choose spanning the path. From an argument exactly parallel to that for themagnetic flux in the third equation (see above), this will be true if andonly If $\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}=0$fora closed surface (with the path lying in the surface—thisclosed surface is made up by combining two different surfaces spanning thepath).

Now, $\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}$ takenover a closed surface is just the net current flow out of the enclosedvolume. Obviously, in a situation with steady currents flowing along wires orthrough conductors, with no charge piling up or draining away from anywhere,this is zero. However, if the total electric charge $q,$ say, enclosed by the closed surface is changingas time goes on, then evidently

$\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}=-dq/dt,$

where we put in a minus sign because, with our convention, $d\stackrel{\to }{A}$ is a little vector pointing outwards,so the integral represents net flow of charge out from the surface,equal to the rate of decrease of the enclosed total charge.

To summarize: if the local charge densities arechanging in time, that is, if charge is piling up in or leaving some region,then $\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}\ne 0$ over a closed surface around that region. Thatimplies that $\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}$ over one surface spanning the wire will be differentfrom $\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}$ over another surface spanning the wire ifthese two surfaces together make up a closed surface enclosing a regioncontaining a changing amount of charge.

The key to fixing this up is to realize that although $\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}=-dq/dt\ne 0,$ the right-hand side can be written as anothersurface integral over the same surface, using the first Maxwellequation, that is, the integral over a closed surface

$\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}=q/{\epsilon }_{0}$

where $q$ is the total charge in the volume enclosed bythe surface.

By taking the time rate of change of both sides, wefind

$\frac{d}{dt}\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}=\frac{1}{{\epsilon }_{0}}\frac{dq}{dt}$

Putting this together with $\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}=-dq/dt$ gives:

${\epsilon }_{0}\frac{d}{dt}\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}+\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}=0$

for any closed surface, and consequently this is asurface integral that must be the same for any surface spanning thepath or circuit! (Because two differentsurfaces spanning the same circuit add up to a closed surface. We’ll ignore thetechnically trickier case where the two surfaces intersect each other, creatingmultiple volumes—there onemust treat each created volume separately to get the signs right.)

Therefore, this is the way to generalize Ampere's law fromthe magnetostatic situation to the case where charge densities are varying withtime, that is to say the path integral

$\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}\int \left(\stackrel{\to }{j}+{\epsilon }_{0}\frac{d\stackrel{\to }{E}}{dt}\right)\cdot d\stackrel{\to }{A}$

and this gives the same result for any surface spanning thepath. A Sheet of Current: ASimple Magnetic Field

As a preliminary to looking at electromagnetic waves, weconsider the magnetic field configuration from a sheet of uniform current oflarge extent. Think of the sheet asperpendicular to this sheet of paper, the current running vertically upwards. It might be helpful to visualize the sheet asmany equal parallel fine wires uniformly spaced close together:

......................................................................................(wires)

The magnetic field from this current sheet can be foundusing Ampere's law applied to a rectangular contour in the plane of the paper,with the current sheet itself bisecting the rectangle, so the rectangle's topand bottom are equidistant from the current sheet in opposite directions. Maxwell's Two Equations For Electrostatic Fields Worksheet

Applying Ampere’s law to the above rectangularcontour, there are contributions to $\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }$ only from the top and bottom, and they add togive $2BL$ if the rectangle has side $L.$ The total current enclosed by the rectangle is$IL,$ taking the current density of the sheet to be $I$ amperes per meter (how many little wires permeter multiplied by the current in each wire).

Thus, $\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}\cdot \text{(enclosed currents)}$ immediately gives:

$B={\mu }_{0}I/2$

a magnetic field strength independent of distance $d$ from the sheet. (This is the magnetostaticanalog of the electrostatic result that the electric field from an infinitesheet of charge is independent of distance from the sheet.) In real life, where there are no infinitesheets of anything, these results are good approximations for distances fromthe sheet small compared with the extent of the sheet.

Switching on the Sheet: How Fast Does the Field Build Up?

Consider now how the magnetic field develops if the currentin the sheet is suddenly switched on at time $t=0.$ We will assume that sufficiently close to thesheet, the magnetic field pattern found above using Ampere's law is ratherrapidly established.

In fact, we will assume further that the magnetic fieldspreads out from the sheet like a tidal wave, moving in both directions at somespeed $v,$ so that after time $t$ the field within distance $vt$ of the sheet is the same as that found abovefor the magnetostatic case, but beyond $vt$ there is at that instant no magnetic fieldpresent.

Let us now apply Maxwell's equations to this guess to see ifit can make sense. Certainly Ampere's law doesn't work by itself, because if wetake a rectangular path as we did in the previous section, for $d everything works as before, but for arectangle extending beyond thespreading magnetic field, $d>vt,$ there will be no magnetic field contribution from the top and bottom of therectangle, and hence

$\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }=0$

but there is definitely enclosed current!

We are forced to conclude that for Maxwell's fourth equationto be correct, there must also be a changing electric field through therectangular contour.

Let us now try to nail down what this electric field throughthe contour must look like. First, itmust be through the contour, that is, have a component perpendicular to theplane of the contour, in other words, perpendicular to the magnetic field. In fact, electric field components in otherdirections won't affect the fourth equation we are trying to satisfy, so weshall ignore them. Notice first that fora rectangular contour with $d Ampere'slaw works, so we don'twant a changing electric fieldthrough such a contour (but a constant electric field would be ok).

Now apply Maxwell's fourth equation to a rectangular contourwith $d>vt,$

It is: path integral $\oint \stackrel{\to }{B}\cdot d\stackrel{\to }{\ell }={\mu }_{0}\int \left(\stackrel{\to }{j}+{\epsilon }_{0}\frac{d\stackrel{\to }{E}}{dt}\right)\cdot d\stackrel{\to }{A}$ (over surface spanning path).

For the rectangle shown above, the integral on the left handside is zero because $\stackrel{\to }{B}$ is perpendicular to $d\stackrel{\to }{\ell }$ along the sides, so the dot product is zero,and $\stackrel{\to }{B}$ is zero at the top and bottom, because theoutward moving 'wave' of magnetic field hasn’t gotten there yet.Therefore, the right hand side of the equation must also be zero.

We know $\int \stackrel{\to }{j}\cdot d\stackrel{\to }{A}=LI$,so we must have: ${\epsilon }_{0}\frac{d}{dt}\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}=-LI.$

Finding the Speed of the Outgoing Field Front: the Connection with Light

So, as long as the outward moving front of magnetic field,travelling at $v,$ hasn't reached the top and bottom of therectangular contour, the electric field through the contour increases linearly with time, but theincrease drops to zero (because Ampere's law is satisfied) the moment the frontreaches the top and bottom of the rectangle. The simplest way to get thisbehavior is to have an electric field of strength $E,$ perpendicular to the magnetic field,everywhere there is a magnetic field, so the electric field also spreadsoutwards at speed $v.$ Notethat, unlike the magnetic field, the electric field must point the same way onboth sides of the current sheet, otherwise its net flux through the rectanglewould be zero.

After time $t,$ then, the electric field flux through therectangular contour $\int \stackrel{\to }{E}\cdot d\stackrel{\to }{A}$ will be just field x area = $E\cdot 2\cdot vtL,$ and the rate of change will be $2EvL.$ (It's spreading both ways, hence the 2).

Therefore

${\epsilon }_{0}E\cdot 2\cdot vtL=-LI,$

the electric field is downwards and of strength $E=I/\left(2{\epsilon }_{0}v\right).$

Since $B={\mu }_{0}I/2,$ this implies:

$B={\mu }_{0}{\epsilon }_{0}vE.$

But we have another equation linking the field strengths ofthe electric and magnetic fields, Maxwell's third equation:

$\oint \stackrel{\to }{E}\cdot d\stackrel{\to }{\ell }=-d/dt\left(\int \stackrel{\to }{B}\cdot d\stackrel{\to }{A}\right)$

We can apply this equation to a rectangular contour withsides parallel to the $E$ field, one side being within $vt$ of the current sheet, the other more distant,so the only contribution to the integral is $EL$ fromthe first side, which we take to have length $L.$ (This contour is all on one side of thecurrent sheet.) The area of the rectangle the magnetic flux is passing throughwill be increasing at a rate $Lv$ (square meters per second) as the magneticfield spreads outwards.

Application Of Electrostatic Field

It follows that

$E=vB.$

Putting this together with the result of the fourthequation,

Electrostatic Field Lines

$B={\mu }_{0}{\epsilon }_{0}vE,$

we deduce

${v}^{2}=\frac{1}{{\mu }_{0}{\epsilon }_{0}}.$

Substituting the defined value of ${\mu }_{0},$ and the experimentally measured value of ${\epsilon }_{0},$ we findthat the electric and magnetic fields spread outwards from the switched-oncurrent sheet at a speed of 3 x 108 meters per second.

Electrostatic Force Equation

This is how Maxwell discovered a speed equal to the speed oflight from a purely theoretical argument based on experimental determinationsof forces between currents in wires and forces between electrostatic charges.This of course led to the realization that light is an electromagnetic wave,and that there must be other such waves with different wavelengths. Hertz detected other waves, of much longerwavelengths, experimentally, and this led directly to radio, tv, radar,etc.