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For individual segments with area density ri and area dAi, the mass ofeach segment is ri dAi, and the mass of the entire surface of N segments is PNgiven byi1 ri dAi. As you can imagine, the smaller you make the areasegments, the closer this gets to the true mass, since your approximationof constant r is more accurate for smaller segments. If you let the seg-ment area dA approach zero and N approach innity, the summationbecomes integration, and you have Z Mass rx; y dA:SThis is the area integral of the scalar function r x, y over the surface S.

Itis simply a way of adding up the contributions of little pieces of afunction the density in this case to nd a total quantity. To understandthe integral form of Gausss law, it is necessary to extend the concept ofthe surface integral to vector elds, and thats the subject of the nextsection.

Density approximately constant over each of these areas dA1, dA2,. Whats a vectoreld? As the name suggests, a vector eld is a distribution of quantities inspace a eld and these quantities have both magnitude and direction,meaning that they are vectors.

So whereas the distribution of temperaturein a room is an example of a scalar eld, the speed and direction of theow of a uid at each point in a stream is an example of a vector eld.

The analogy of uid ow is very helpful in understanding the meaningof the ux of a vector eld, even when the vector eld is static andnothing is actually owing. You can think of the ux of a vector eldover a surface as the amount of that eld that ows through thatsurface, as illustrated in Figure 1. To work those kinds of problems, youll need tounderstand how to extend the concept of the surface integral to vectorelds.

Before proceeding, you should think for a moment about how youmight go about nding the rate of ow of material through surface S. You can dene rate of ow in a few different ways, but it will help toframe the question as How many particles pass through the membraneeach second?

But look again at Figure 1. The different lengths of the arrows aremeant to suggest that the ow of material is not spatially uniform,meaning that the speed may be higher or lower at various locationswithin the ow.

This fact alone would mean that material ows throughsome portions of the surface at a higher rate than other portions, but youmust also consider the angle of the surface to the direction of ow. Anyportion of the surface lying precisely along the direction of ow willnecessarily have zero particles per second passing through it, since theow lines must penetrate the surface to carry particles from one side to Thus, you must be concerned not only with the speed of owand the area of each portion of the membrane, but also with the com-ponent of the ow perpendicular to the surface.

AE But does the analogy used in theprevious section mean that the electric ux should be thought of as a owof particles, and that the electric eld is the product of a density and avelocity? The answer to this question is absolutely not. Remember that whenyou employ a physical analogy, youre hoping to learn something aboutthe relationships between quantities, not about the quantities themselves.

So, you can nd the electric ux by integrating the normal component ofthe electric eld over a surface, but you should not think of the electricux as the physical movement of particles. How should you think of electric ux? One helpful approach followsdirectly from the use of eld lines to represent the electric eld. Recallthat in such representations the strength of the electric eld at any point isindicated by the spacing of the eld lines at that location.

More specif-ically, the electric eld strength can be considered to be proportional tothe density of eld lines the number of eld lines per square meter in aplane perpendicular to the eld at the point under consideration.

Inte-grating that density over the entire surface gives the number of eld linespenetrating the surface, and that is exactly what the expression forelectric ux gives. Thus, another way to dene electric ux is electric flux UE number of field lines penetrating surface.

There are two caveats you should keep in mind when you think of electricux as the number of electric eld lines penetrating a surface. The rst isthat eld lines are only a convenient representation of the electric eld,which is actually continuous in space. The number of eld lines you Thus, asurface penetrated by ve eld lines in one direction say from the topside to the bottom side and ve eld lines in the opposite direction frombottom to top has zero ux, because the contributions from the twogroups of eld lines cancel.

So, you should think of electric ux as the netnumber of eld lines penetrating the surface, with direction of penetra-tion taken into account. If you give some thought to this last point, you may come to animportant conclusion about closed surfaces. Consider the three boxesshown in Figure 1. The box in Figure 1.

Thus,every eld line that enters must leave, and the ux through the box mustbe zero. En Now consider the box in Figure 1. The surfaces of this box arepenetrated not only by the eld lines originating outside the box, but alsoby a group of eld lines that originate within the box. In this case, the netnumber of eld lines is clearly not zero, since the positive ux of the lines Gausss law for electric elds 15that originate in the box is not compensated by any incoming negative ux.

Thus, you can say with certainty that if the ux through any closedsurface is positive, that surface must contain a source of eld lines. Finally, consider the box in Figure 1. In this case, some of the eldlines terminate within the box. These lines provide a negative ux at thesurface through which they enter, and since they dont exit the box, theircontribution to the net ux is not compensated by any positive ux.

Clearly, if the ux through a closed surface is negative, that surface mustcontain a sink of eld lines sometimes referred to as a drain. Now recall the rst rule of thumb for drawing charge-induced electriceld lines; they must originate on positive charge and terminate onnegative charge.

So, the point from which the eld lines diverge in Figure1. If the amount of charge at these locations were greater, there would bemore eld lines beginning or ending on these points, and the ux throughthe surface would be greater.

And if there were equal amounts of positiveand negative charge within one of these boxes, the positive outward uxproduced by the positive charge would exactly cancel the negative inward ux produced by the negative charge.

So, in this case the uxwould be zero, just as the net charge contained within the box would bezero. You should now see the physical reasoning behind Gausss law: theelectric ux passing through any closed surface that is, the number ofelectric eld lines penetrating that surface must be proportional to thetotal charge contained within that surface.

Before putting this concept touse, you should take a look at the right side of Gausss law. Simply put, it is because any charge located out-side the surface produces an equal amount of inward negative ux andoutward positive ux, so the net contribution to the ux through thesurface must be zero.

How can you determine the charge enclosed by a surface? In someproblems, youre free to choose a surface that surrounds a knownamount of charge, as in the situations shown in Figure 1. In each ofthese cases, the total charge within the selected surface can be easilydetermined from geometric considerations.

For problems involving groups of discrete charges enclosed by surfacesof any shape, nding the total charge is simply a matter of adding theindividual charges. X Total enclosed charge qi : iWhile small numbers of discrete charges may appear in physics andengineering problems, in the real world youre far more likely to encountercharged objects containing billions of charge carriers lined along a wire,slathered over a surface, or arrayed throughout a volume.

In such cases,counting the individual charges is not practical but you can determinethe total charge if you know the charge density. Charge density may bespecied in one, two, or three dimensions 1-, 2-, or 3-D. Point ChargedMultiple charge lineChargedpoint charges planeEnclosing enclosing pillbox Enclosing Enclosing cubespherecylinder Figure 1. In such cases,the integration techniques described in the Surface Integral section ofthis chapter must be used.

Thus,Z 1-D : qenc k dl where k varies along a line; LZ2-D : qenc r da where r varies over a surface; S Z3-D : qenc q dV where q varies over a volume: VYou should note that the enclosed charge in Gausss law for electric eldsis the total charge, including both free and bound charge. You can readabout bound charge in the next section, and youll nd a version ofGausss law that depends only on free charge in the Appendix.

Once youve determined the charge enclosed by a surface of any sizeand shape, it is very easy to nd the ux through that surface; simplydivide the enclosed charge by e0, the permittivity of free space. Thephysical meaning of that parameter is described in the next section. The permittivity of a material determinesits response to an applied electric eld in nonconducting materials called insulators or dielectrics , charges do not move freely, butmay be slightly displaced from their equilibrium positions.

The relevantpermittivity in Gausss law for electric elds is the permittivity of freespace or vacuum permittivity , which is why it carries the subscriptzero. The value of the vacuum permittivity in SI units is approximately8.

A more precise value for the permittivity of freespace ise0 8. Does the presence of this quantity mean that this form of Gausss law isonly valid in a vacuum?

No, Gausss law as written in this chapter isgeneral, and applies to electric elds within dielectrics as well as those infree space, provided that you account for all of the enclosed charge,including charges that are bound to the atoms of the material. The effect of bound charges can be understood by considering whathappens when a dielectric is placed in an external electric eld.

Inside thedielectric material, the amplitude of the total electric eld is generally lessthan the amplitude of the applied eld. The reason for this is that dielectrics become polarized when placedin an electric eld, which means that positive and negative charges aredisplaced from their original positions.

And since positive charges aredisplaced in one direction parallel to the applied electric eld andnegative charges are displaced in the opposite direction antiparallel tothe applied eld , these displaced charges give rise to their own electriceld that opposes the external eld, as shown in Figure 1. This makesthe net eld within the dielectric less than the external eld.

It is the ability of dielectric materials to reduce the amplitude of anelectric eld that leads to their most common application: increasing thecapacitance and maximum operating voltage of capacitors.

As youmay recall, the capacitance ability to store charge of a parallel-platecapacitor is High-permittivity materials canprovide increased capacitance without requiring larger plate area ordecreased plate spacing. The relative permittivity of ice, for example,changes from approximately 81 at frequencies below 1 kHz to less than 5 atfrequencies above 1 MHz.

Most often, it is the low-frequency value ofpermittivity that is called the dielectric constant. One more note about permittivity; as youll see in Chapter 5, thepermittivity of a medium is a fundamental parameter in determining thespeed with which an electromagnetic wave propagates through thatmedium. At this point, you should be convinced that Gausss law relatesthe electric ux through a closed surface to the charge enclosed by thatsurface. Here are some examples of what can you actually do with thatinformation.

Example 1. Problem: Five point charges are enclosed in a cylindrical surface S. Gausss law for electric elds21Example 1. If the ux through the surface of thesphere is 1. Problem: A point source of charge q is placed at the center of curvature ofa spherical section that extends from spherical angle h1 to h2 and from u1to u2.

Find the electric ux through the spherical section. Solution: Since the surface of interest in this problem is open, youll haveto nd the electric ux by integrating the normal component of theelectric eld over the surface. You can then check your answer usingGausss law by allowing the spherical section to form a complete spherethat encloses the point charge.

From Table 1. Gausss law for electric elds23 Z Z ZZ1 q 2qUE r sin h dh df sin h dh df;hf 4pe0 r 2 4pe0 hfwhich is easily integrated to give qUE cos h1 cos h2 f2 f1 :4pe0As a check on this result, take the entire sphere as the section h1 0,h2 p, u1 0, and u2 2p.

This givesqqUE 1 1 2p 0 ; 4pe0e0exactly as predicted by Gausss law. Problem: The electric eld at distance r from an innite line charge withlinear charge density k is given in Table 1. After all, while the electric eld does appear in the equation, it is onlythe normal component that emerges from the dot product, and it is onlythe integral of that normal component over the entire surface that is propo-rtional to the enclosed charge. Do realistic situations exist in which it ispossible to dig the electric eld out of its interior position in Gausss law?

Happily, the answer is yes; you may indeed nd the electric eldusing Gausss law, albeit only in situations characterized by high sym-metry. Specically, you can determine the electric eld whenever youreable to design a real or imaginary special Gaussian surface thatencloses a known amount of charge.

A special Gaussian surface is one onwhich 1 the electric eld is either parallel or perpendicular to the surfacenormal which allows you to convert the dot product into analgebraic multiplication , and 2 the electric eld is constant or zero over sections of the surface whichallows you to remove the electric eld from the integral.

Of course, the electric eld on any surface that you can imagine aroundarbitrarily shaped charge distributions will not satisfy either of theserequirements. But there are situations in which the distribution of chargeis sufciently symmetric that a special Gaussian surface may be imagined. Specically, the electric eld in the vicinity of spherical charge distribu-tions, innite lines of charge, and innite planes of charge may bedetermined by direct application of the integral form of Gausss law.

Geometries that approximate these ideal conditions, or can be approxi-mated by combinations of them, may also be attacked using Gausss law.

The following problem shows how to use Gausss law to nd theelectric eld around a spherical distribution of charge; the other cases arecovered in the problem set, for which solutions are available on thewebsite. Solution: Consider rst the electric eld outside the sphere. Since thedistribution of charge is spherically symmetric, it is reasonable to expectthe electric eld to be entirely radial that is, pointed toward or awayfrom the sphere.

But the charge is uniformly distrib-uted throughout the sphere, so there can be no preferred direction ororientation rotating the sphere simply replaces one chunk of chargewith another, identical chunk so this can have no effect whatsoever onthe electric eld.

Faced with this conundrum, you are forced to concludethat the electric eld of a spherically symmetric charge distribution mustbe entirely radial.

For a Eradial electric eld, there can be only one choice; your Gaussian surfacemust be a sphere centered on the charged sphere, as shown in Figure 1.

Notice that no actual surface need be present, and the special Gaussiansurface may be purely imaginary it is simply a construct that allows youto evaluate the dot product and remove the electric eld from the surfaceintegral in Gausss law. You can now useGausss law to nd the value of the electric eld: Gausss law for electric elds 27 nSpecialGaussiansurfaceRadialelectricfieldn Chargedn sphere n Figure 1. You can usethis expression to nd the electric eld both outside and inside the sphere.

To nd the electric eld outside the sphere, construct your Gaussiansurface with radius ra so that the entire charged sphere is within theGaussian surface. In this case, the enclosed charge is the charge Gausss law for electric elds For now, make sure you grasp the main idea of Gausss law in differentialform:The electric eld produced by electric charge diverges from positivecharge and converges upon negative charge.

In other words, the only places at which the divergence of the electric eldis not zero are those locations at which charge is present. If positivecharge is present, the divergence is positive, meaning that the electric eldtends to ow away from that location. If negative charge is present, thedivergence is negative, and the eld lines tend to ow toward thatpoint. Note that theres a fundamental difference between the differentialand the integral form of Gausss law; the differential form deals with thedivergence of the electric eld and the charge density at individual pointsin space, whereas the integral form entails the integral of the normalcomponent of the electric eld over a surface.

Familiarity with both formswill allow you to use whichever is better suited to the problem youretrying to solve. In any problem inwhich the spatial variation of the vector electric eld is known at aspecied location, you can nd the volume charge density at that locationusing this form.

And if the volume charge density is known, thedivergence of the electric eld may be determined. This symbol represents a vector differential operatorcalled nabla or del, and its presence instructs you to take derivativesof the quantity on which the operator is acting. Each of these operations is discussed in later sections; for now well justconsider what an operator is and how the del operator can be written inCartesian coordinates.

Like all good mathematical operators, del is an action waiting tohappen. This expression may appear strange, since in thisform it is lacking anything on which it can operate. James Clerk Maxwell coined the term convergence to describe themathematical operation that measures the rate at which electric eld linesow toward points of negative electric charge meaning that positiveconvergence was associated with negative charge. A few years later,Oliver Heaviside suggested the use of the term divergence for the samequantity with the opposite sign.

Thus, positive divergence is associatedwith the ow of electric eld lines away from positive charge. Both ux and divergence deal with the ow of a vector eld, butwith an important difference; ux is dened over an area, while diver-gence applies to individual points.

In the case of uid ow, the divergenceat any point is a measure of the tendency of the ow vectors to divergefrom that point that is, to carry more material away from it than isbrought toward it. Thus points of positive divergence are sources fau-cets in situations involving uid ow, positive electric charge in electro-statics , while points of negative divergence are sinks drains in uid ow,negative charge in electrostatics. The mathematical denition of divergence may be understood byconsidering the ux through an innitesimal surface surrounding thepoint of interest.

Youll nd a more user-friendly mathematical expression fordivergence later in this section, but rst you should take a look at thevector elds shown in Figure 1. To nd the locations of positive divergence in each of these elds, lookfor points at which the ow vectors either spread out or are largerpointing away from the location and shorter pointing toward it.

Someauthors suggest that you imagine sprinkling sawdust on owing water toassess the divergence; if the sawdust is dispersed, you have selected apoint of positive divergence, while if it becomes more concentrated,youve picked a location of negative divergence. Gausss law for electric elds 33 a b c Figure 1. Using such tests, it is clear that locations such as 1 and 2 in Figure 1. The divergence at various points in Figure 1.

Location 5 is obviously a point of positive divergence, but what aboutlocations 6 and 7? The ow lines are clearly spreading out at those loca-tions, but theyre also getting shorter at greater distance from the center. Does the spreading out compensate for the slowing down of the ow? Answering that question requires a useful mathematical form of thedivergence as well as a description of how the vector eld varies fromplace to place. Note thatthe divergence of a vector eld is a scalar quantity; it has magnitude butno direction.

In Figure1. This expression is positive for 0x1, 0 at 2x 1, and negative for 1x3, just as your visual inspection suggested. Finally, consider the vector eld in Figure 1. The ow lines are spreadingout as they were in Figure 1. As you consider the divergence of the electric eld, you shouldremember that some problems may be solved more easily using non-Cartesian coordinate systems.

In electrostatics,all electric eld lines begin on points of positive charge and terminateon points of negative charge, so it is understandable that this expressionis proportional to the electric charge density at the location underconsideration. Consider the electric eld of the positive point charge; the electric eldlines originate on the positive charge, and you know from Table 1.

The reason the origin where r 0 is not included in the previousanalysis is that the expression for the divergence includes terms con-taining r in the denominator, and those terms become problematic as rapproaches zero. Gausss law for electric elds 37 It is worth your time to make sure you understand the signicance ofthis last point. A casual glance at the electric eld lines in the vicinity of apoint charge suggests that they diverge everywhere in the sense ofgetting farther apart.

The key factor in determining the divergence at anypoint is not simply the spacing of the eld lines at that point, but whetherthe ux out of an innitesimally small volume around the point is greaterthan, equal to, or less than the ux into that volume. If the outward uxexceeds the inward ux, the divergence is positive at that point. If theoutward ux is less than the inward ux, the divergence is negative, and ifthe outward and inward uxes are equal the divergence is zero at thatpoint.

In the case of a point charge at the origin, the ux through an inni-tesimally small surface is nonzero only if that surface contains the pointcharge. Everywhere else, the ux into and out of that tiny surface must bethe same since it contains no charge , and the divergence of the electriceld must be zero. The following examples should help you understand how to solveproblems of this type. Problem: If the vector eld of Figure 1. Solution: When confronted with a problem like this, you may be temptedto dive in and immediately begin taking derivatives to determine thedivergence of the eld.

A better approach is to think about the eld for amoment and to attempt to visualize the eld lines a task that may bedifcult in some cases.

Fortunately, there exist a variety of computationaltools such as MATLAB and its freeware cousin Octave that areimmensely helpful in revealing the details of a vector eld. If youre surprised by the direction of the eld, consider that thex-component of the eld depends on y so the eld points to the rightabove the x-axis and to the left below the x-axis , while the y-componentof the eld depends on the negative of x so the eld points up on the leftof the y-axis and down on the right of the y-axis.

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