Lesson 7 Gauss s Law and Electric Fields

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1 Lesson 7 Gauss s Law and Electic Fields Lawence B. Rees 7. You may make a single copy of this document fo pesonal use without witten pemission. 7. Intoduction While it is impotant to gain a solid conceptual foundation of field lines and contous, it is also impotant to undestand electicity and magnetism with mathematical pecision. In this chapte, we will tanslate concepts such as numbe of field lines into mathematical functions and use these esults to find the electic fields of seveal impotant distibutions of chage. This will be ou culminating wok with electostatics. 7.1 Gauss's Law of Electicity Let us biefly eview a few things about electic fields and electic field lines. Expeimentally we know that the Coulomb foce between chaged paticles falls off as 1 /. In Lesson 5 we saw adially diected electic field lines natually cay this dependence as long we take the field stength to be given by the numbe of lines pe unit aea passing though a section of a pependicula suface. Since we choose the numbe of field lines going towad a positive point chage o away fom a negative point chage to be popotional to the magnitude of the chage, electic field lines geometically give us the same infomation that Coulomb s law gives us algebaically: F q /. When thee ae moe complicated aangements of point chages, the nea field and the fa field both esemble the fields of single point chages. We have dawn the field lines fo seveal aangements of chages, so we know how moe complicated field lines behave. Think About It A positive point chage alone in space has field lines coming unifomly outwad. As a negative chage is bought neae to it, what happens to the field lines? Does it make any diffeence what the magnitude of the second chage is? What happens when a positive chage is bought nea the fist chage? Keeping these facts in mind, let us now think of a box with vaious unseen chages inside. We ae given the task of tying to detemine as much as we can about the chages in the box fom the electic field lines which extend outside the box. Fo sake of agument, let us suggest that we daw fou field lines fo each unit of chage, as we did in Lesson 5. Think About It What can we say about the numbe of field lines coming out the box o going into the box if a chage of + is inside the box? if a chage of 3 is inside the box? if both chages ae located vey close to each othe inside the box? if both chages ae located on opposite ends of the box? 1

2 As you pobably deduced, we cannot completely detemine the aangement of chage inside the box by counting the numbe of field lines outside. A simple chage of +1 will poduce almost the same field outside the box as a chage of +187 and a chage of 186 located nea each othe. Futhemoe, if chages +1 and 1 ae both located inside the box, some field lines fom the positive chage may go diectly to the negative chage without eve leaving the box, and some field lines may leave the box fom the positive chage and come back in to the negative chage. The one thing we can conclude, howeve, is that the diffeence between the total numbe of field lines leaving the box and the total numbe of field lines enteing the box is fou times the net chage within the box. Let me belabo this point a little, just in case the conclusion is not yet obvious. We call let the net numbe of field lines passing though the box be defined as the numbe of field lines passing out of the box minus the numbe of field lines enteing the box. If thee is no chage inside the box but chages located nea the box on the outside, each field line that passes into the box passes out of the box again, leaving the net numbe of field lines equal to zeo. If one unit of positive chage is in the box, the net numbe of field lines is +4 (4 because we abitaily chose fou lines pe unit of chage). If one unit of negative chage is in the box, the net numbe of field lines is 4. If chages +1 and 1 ae both in the box, the field line fom the positive chage could do one of two things: 1) they could go to the negative chage without passing out of the box so thee would be no contibution to the net numbe of field lines, o ) they could leave the box and come back in to the negative chage so each line would contibute +1 to the net numbe of field lines as it passes out of the box and 1 to the net numbe as it comes back in. The final esult is that the net numbe of field lines is. Continuing this agument, we may conclude that net numbe of field lines 4 the total chage inside. This conclusion, of couse, depends neithe on the shape no size of the box. Any closed suface which contains a volume is adequate. Such sufaces ae called "Gaussian sufaces." You may safely think of a Gaussian suface as a balloon of any abitay shape. Using these definitions of "net numbe" and "Gaussian suface," we may wite the following genealization: Gauss's Law of Electicity The net numbe of electic field lines passing though a Gaussian suface is popotional to the total chage enclosed by the Gaussian suface. Gauss's law of electicity is the fist of Maxwell's equations. Often we efe to it simply as "Gauss's law." The concept is simple pehaps so simple that it seems useless. Howeve, it foms one of the conestones of electomagnetic theoy. In the end we must emembe that Gauss's law is a diect consequence of the way we constuct electic field lines. And electic field lines in tun ae a geometic way of epesenting Coulomb's law. Schematically we may wite: Gauss's law 6 electic field lines 6 Coulomb's law. In othe wods, Gauss's law is nothing moe than a diffeent way of expessing Coulomb's law. Howeve, because it is a diffeent way of expessing the same physical content, it can be useful

3 ; to in a completely diffeent set of applications. Late we will find mathematical ways of expessing this concept and use it quantitatively in detemining the electic field of seveal objects. Fo now, we will find that applying it conceptually can lead to some inteesting and useful esults. Things to emembe: A Gaussian suface is a closed suface a suface that encloses a volume. Gauss's law of electicity: The net numbe of electic field lines passing though a Gaussian suface is popotional to the total chage enclosed by the Gaussian suface. Gauss s law is equivalent to Coulomb s law because it just an obsevation based on field lines, and field lines ae geometical epesentations of Coulomb s law. 7. Chage Density Befoe we apply Gauss's law, howeve, we need to define chage density a little moe caefully than we did ealie. Actually, we need to define thee sepaate ideas collectively known as chage density. When you think of density, you pobably think of the mass of a cetain amount of mateial. ρ The most impotant chage density is "volume chage density." You ae pobably used to defining mass density as mass pe unit volume. Volume chage density is defined analogously to this. Fo example, a cubic mete of wate weighs 1 kg, so the density of wate is 1 kg/m 3. (O since these ae athe lage quantities, we usually expess the density of wate as 1 g/cm 3.) We use the Geek lette (ho) to denote volume chage density. Thus chage / unit volume. You might ask why we should woy about the density of chage at all. Can t we just expess eveything in tems of the dimensions of an object and its total chage? If the chage on the object is unifom, sometimes this is tue; howeve, in many objects the chage is not unifomly distibuted thoughout. In these cases we need a way of expessing just whee the chage is located. If q is the chage contained in a small volume v at Catesian coodinates x, y, and z, we may wite the volume density as function of position in the following fashion: q ρ ( x, y, z). v ρ That is, the volume chage density is the chage pe unit volume fo a small volume located at a specified point in space. In ode fo be a "good" function, we eally must take the limit of the above expession as v howeve, this distinction usually is not of significance to us. (Note: In this text we shall use a lowe case v fo volume to pevent confusion with electic potential, V.) 3

4 (lambda). (sigma). Think About It Descibe in wods the following chage densities: 6, if ρ(, θ, φ), othewise 3xy, if x < 1, y < 1, z < 1, ρ( x, y, z), othewise σ Wheeas any massive object will always take up volume, electic chage can be confined to the suface of an object. Since the suface laye is, fo all pactical puposes, infinitely thin, it becomes impossible to define a volume chage density fo the suface. Fo this type of situation, we define a "suface chage density" as the chage pe unit aea on a suface. We denote suface chage density by the Geek lette Think About It Descibe in wods the following chage densities: 3 σ ( x, y) xy cos( θ ), if θ < π σ ( θ, φ) cos( θ ), if θ < π λ Also, chage may be located along a staight line. In that case, we can define the "linea chage density" as the chage pe unit length along the line. The linea chage density is epesented by the Geek lette Things to emembe; Chage densities may be volume chage densities (D chage/volume), suface chage density (F chage/aea), o linea chae density (8 chage/unit length). 7.3 The Fields of Chage Distibutions with Radial Spheical Symmety Conside a sphee with a chage distibution having adial symmety. Recall that by ou definition of adial spheical symmety, this means the volume chage density may vay with 4

5 distance fom the oigin of a coodinate system,, but not with angle. We agued ealie that, because of the symmety of the poblem, the electic field lines outside the sphee must be diected adially outwad fom the oigin and be evenly spaced fom one anothe. Because of this, we concluded that the electic field must be the same as the field of a point chage having the same total chage as the sphee. But what can we conclude about the electic field within the sphee? Fist let us conside a spheical shell with a unifom suface chage density. (This is just a shell, like a balloon o the ubbe pat of a basketball, with no chage at all inside.) To find the electic field inside the shell we fist constuct a Gaussian suface within it. Let the Gaussian suface be a sphee of adius, whee is smalle than the adius of the chaged shell. We ll place the Gaussian suface inside the shell and concentic with it. We know that the net numbe of field lines passing though this Gaussian suface is popotional to the chage enclosed within the Gaussian suface. But the enclosed chage is zeo, so the net numbe of field lines passing though the suface is also zeo! We might be tempted to immediately conclude that the electic field is zeo inside the shell; howeve, we must be caeful hee because all Gauss's law tells us is that the net numbe of field lines passing though the suface is zeo. This means that the numbe of field lines enteing the sphee must be the same as the numbe leaving. Fotunately, we can apply symmety aguments once again. Since the chage distibution is adially symmetic, we know that the electic field lines within the sphee must all be diected eithe outwad o inwad. (They cannot lean to one side o anothe as no side is diffeent fom the othes. They cannot be patly outwad and patly inwad as no pat of the spheical shell is diffeent fom the est.) If this is tue, the only way we can have no net field lines passing though the suface is to have no field lines at all, and hence thee is no electic field inside. Figue 7.1. A Gaussian suface inside a chaged basketball. It is impotant to notice how we used Gauss s Law in conjunction with symmety to aive at this conclusion! 5

6 Figue 7.. Electic field lines that ae impossible because they don t have adial symmety. Think About It Coulomb's Law tells us that the close we ae to a point souce, the stonge the field must be. Why is it then that if we put a small positive test chage just inside a spheical shell, it isn't epelled towad the cente of the shell? Hint: You need to think about all of the chage, not just the closest chage. We can easily see that the same conclusion must apply if the shell is thin o thick. As long as the chage distibution has adial spheical symmety, the electic field within a hollow sphee must be zeo. We can also apply Gauss's law along with symmety aguments to detemine the electic field within a solid sphee having a adially symmetic chage density. By symmety, we know that the electic field lines ae diected adially outwad and ae evenly spaced. Let us constuct a Gaussian suface at a adius within such a sphee. The numbe of field lines passing out of this suface is popotional to the chage enclosed by the suface. Hence the electic field is the same as the electic field of a point chage having the same magnitude as the faction of total chage enclosed within a sphee of adius. All we have to do is use geomety, pehaps with integation, to detemine the enclosed chage. Wheneve thee is adial spheical symmety (not necessaily othewise!), the chage outside the Gaussian suface contibutes nothing to the electic field. Mathematically: Electic field inside a spheically symmetic chage distibution 1 Qenclosed (7.1) E. 4πε 6

7 Things to emembe: We can use Gauss s law with symmety to calculate the electic field of spheically symmetic chage distibutions. Know how to do that! Outside the chage distibution, the field is the same as the field of a point chage with the same total chage. Inside a hollow sphee with a spheically symmetic chage density, the electic field is zeo. At a adius inside the distibution, the field is the same as the field of a point chage that has the same chage as the chage inside a Gaussian suface of adius. 7.4 Gauss s Law and Conductos One othe impotant application of Gauss's law is to electical conductos. In Lesson 3, we aived at seveal ules fo conductos. Let s fist eview two of those ules: The static electic field eveywhee inside a conducto is zeo. The static electic field on the suface of a conducto is eveywhee pependicula to the suface. Now let us apply Gauss's law. Fist we constuct an abitaily shaped Gaussian suface which lies completely within the conducto. Since thee is no electic field in the conducto, the net numbe of electic field lines passing though the suface is zeo. By Gauss's law, this implies that the total chage contained within the sufaces is also zeo. We can make tiny Gaussian sufaces o lage Gaussian sufaces. We can even make a Gaussian suface just baely inside the outside suface of the entie conducto. The total chage in evey such suface is zeo. In fact, the only place we could find any net chage is on the outside suface of the conducto. Hence, Gauss's law gives us the next impotant ule fo conductos: In a static situation, all net chage on a solid conducto must lie on its oute suface. We agued that this esult was tue in Lesson 3, but we didn t have the tools to constuct a igoous poof at that time. Figue 7.3. Any Gaussian suface within a conducto contains zeo net chage. 7

8 Think About It What can you say about the chage in a conducto if thee is a hollow bubble inside the conducto? What if thee is a chage within a hollow bubble inside a conducto? If thee is a bubble within a conducto, we can place a Gaussian suface just inside the conducto aound the bubble. Since E eveywhee on the Gaussian suface, no net field lines pass though the suface, so the total enclosed chage must be zeo. That doesn t peclude some positive suface chage on one side of the bubble and some negative suface chage on the othe side; howeve, if thee ae no fields to keep the chage sepaated, chage will flow until the suface of the bubble is neutal eveywhee. On the othe hand, if thee is a positive chage in the middle of the bubble, thee must be a negative suface chage on the bubble so that the total chage inside the Gaussian suface emains zeo. We see that the chage in a conducto can and must aange itself on the conducto s suface in such a fashion that it sets up a field of its own which 1) is equal and opposite to the extenal electic field eveywhee inside the conducto and ) is equal and opposite the component of the electic field paallel to the suface of the conducto eveywhee on the suface. Things to emembe: In a static situation, all net chage on a conducto must lie on its oute suface. Know how to pove that this is tue using Guass s law in conjunction with the fact that the static electic field inside a conducto is zeo. 7.5 Gauss's Law of Magnetism If such a thing as magnetic chage wee to exist, and magnetic chage obeyed a 1 / type foce law, as we would be expect the magnetic field to obey a law much like Gauss's law of electicity. That is, the net numbe of magnetic field lines passing though a Gaussian suface must be popotional to the magnetic chage enclosed within the suface. Since thee ae no known magnetic chages, Gauss's law of magnetism is: Gauss's Law of Magnetism The net numbe of magnetic field lines passing though a Gaussian suface is always zeo. This implies that fo evey possible Gaussian suface we can constuct, wheneve a magnetic field line entes the suface, it has to pass out of the suface as well. The only way this can happen is if evey magnetic field line foms a closed loop. This is a vey impotant esult you need to emembe! 8

9 Gauss's law of magnetism is the second of Maxwell's equations. It is equivalent to the expeimental obsevation that thee ae no magnetic monopoles a magnet can t have a noth pole without a south pole. Things to emembe: Magnetic field lines always fom closed loops. Gauss's Law of Magnetism: the net numbe of magnetic field lines passing though a Gaussian suface is always zeo. 7.6 Unifom and Non-unifom Densities In Section 7.3, we found the electic field inside o outside any spheically symmetic 1 Q( inside adius ) chage distibution is given by Eq. (7.1): E( ). All we need to know 4πε to solve the poblem is the total chage within a spheical Gaussian suface of adius. Example 7.1. Unifom Spheical Chage Distibution What is the electic field inside a sphee of adius a having total chage Q if the chage distibution within the sphee is unifom? We need to find how much of the sphee s chage is located within a adius to solve this poblem. Since the chage density is unifom, we can use the elationship Q V enclosed ρ. enclosed This elationship holds equally well at adius a and adius. Thus: Q Qenclosed ρ π a π 3 3 Q enclosed Q 1 E( < a) Q 4πε a 3 a Q 3 4πε a Notice that the electic field inside the sphee inceases linealy with adius. This is because the 3 volume of chage that contibutes to the electic field is popotional to while the Coulomb foce dops off as 1 /. 9

10 If the density vaies with position, howeve, finding the total chage is not so simple. Let us again conside a volume chage density. Even if the density vaies ove a sphee, the density in one vey small egion within the sphee is nealy constant. Thus the chage q within a small volume, V, located at the position is just q ρ 1 ( ) V If we want to know the total chage on the object, we have to divide the entie sphee into small elements of volume. Each volume element must have a density that is (nealy) constant. We then must add up all the contibutions fom each element to obtain the total chage. That is: q dq ρ V. Of couse, this expession is only appoximate unless we let the size of each volume element shink to zeo. In this limit, the sum becomes an integal and q dq ρ ( ) dv. Simila elationships apply to suface chage densities and linea chage densities: ( ) q dq σ, λ ( ) da q dq ( x)dx Often we also need to know the total volume o aea of a egion. To find the total volume, fo example, all we need to do is add up all the elemental volumes: V dv. Sometimes students panic at the mention of integals; howeve, thee ae thee things you must keep in mind: 1) An integal is conceptually nothing moe than a sum. In fact, the symbol is just the lette "s" (fo sum) as used in pinting fonts until about 18. ) Integals ae often much easie to evaluate than the coesponding summations. 3) Integals ae tabulated in tables so we usually do not have to esot to the cleve methods you leaned in calculus classes to evaluate them. Things to emembe: You can use Coulomb s law to find the electic field at adius fo any spheically symmetic chage distibution as long as you emembe that the chage that contibutes to the field is only the chage on the inside of a sphee of adius. In geneal, you can find the enclosed chage by integating ove the chage density: q ρ dv. 7.7 Two-dimensional Integation Most of the integals we will need to evaluate ae in eithe two o thee dimensions. You pobably do not feel vey confident about evaluating such integals; howeve, if you can.

11 visualize what you ae integating, you can always set up the coect integal. To illustate this point, let us begin with integation ove sufaces. Fist of all, let us conside a ectangle with cones (,), (b,), (b,c), and (,c) in Catesian coodinates, as shown below. y c b x Figue 7.4 The standad ectangle Let s find the total aea of the ectangle. Of couse, we know A bc; howeve, we will find A by integation. All we have to do is add up the elements of aea to get the whole. Thus A da. What we do now is the most difficult pat of any integation. We have to slice up the egion; that is, we have to decide how to choose the elemental aeas da. At this point it is impotant to emembe that an integal is just a sum. Any way in which we can daw small aeas and can find da fo each of these aeas is acceptable. In calculus classes you pobably leaned to divide such a ectangle into small ectangles. We daw one abitay ectangle in the lage ectangle, as shown in Fig We denote the length of the sides by dx and dy. Then we can see that da dx dy. To finish the integation, we need to find the limits of integation. These clealy extend in the x diection fom to b and in the y diection fom to c. Thus: A c b da dxdy bc 11

12 y c dy dx b x Figue 7.5 Slicing the ectangle in both x and y diections. Thee s nothing wong with slicing the ectangle into dx dy aeas; howeve, it is often possible to educe a two-dimensional o thee-dimensional integal into a one-dimensional integal by looking at the details of the poblem. In this poblem, we could slice the ectangle along the x diection as shown in Fig. 7.6, o along the y diection as shown in Fig In these cases the aeas ae da c dx and da b dy. The integals ae then evaluated by simply adding up the das: A A da da b c c dx bc b dy bc. o 1

13 y c dx b x Figue 7.6 Slicing the ectangle along the x diection. y c dy b x Figue 7.7 Slicing the ectangle along the y diection. The answe is the same no matte how we do it, but the integal is easie to evaluate when we educe it to a one-dimensional integal. 13

14 da must on Think About It How many othe ways can you slice the ectangle into small aeas da? What is the value fo da in each case? Of couse, this case was tivial as we knew the answe fom the beginning. Let us do a moe complicated poblem now. We take the same ectangle as in Fig. 7.4; howeve, let us find the total chage on the ectangle when the suface chage density is given by the function σ(x) Kx, whee K is a constant. We then slice the ectangle into small units of aea da with chage σ(x) da. We know that the total chage is just q dq σ da. Now let's look at how we could slice the ectangle. If we slice the ectangle in both the x and y diections, the aea is da dx dy and the chage on the slice is dq K x dx dy. We could then evaluate the integal using the same limits as above. The only diffeence is that the integand is now a function of x. c b q K 1 x dx dy Kcb 3 σ σ σ If we slice the ectangle along the y axis, we wite da b dy fo the aea and dq σ(x) b dy fo the chage on this aea. But thee is one poblem. What is the value of this slice? Since the chage density vaies with x, the slice has much moe chage on the ight side than on the left. When we wite dq fo a slice, be a constant on the slice! That is, the simple expession (chage) (suface chage density) (aea) is only valid when the density is constant. We can daw an extemely impotant conclusion fom this: 3 Slicing Rule You must integate ove all vaiables which appea in the integand, being caeful not to ignoe vaiables "hidden" in functions. Hee, the integand has an x appeaing in it, so we must integate ove dx. Slicing the egion into smalle ectangles, as in Fig. 7.5, is fine because we do integate ove dx. Howeve, since y does not appea in the integand, we don t need to integate ove dy. This means we can find a simple way to slice the ectangle so we only integate ove dx. The clea choice is to slice the ectangle as in Fig Hee da c dx and dq Kx c dx. Thus, we find the chage by integating ove dx: b 1 3 q dq Kc x dx Kcb. 3 14

15 Be sue that you thooughly undestand the above examples. If you do, then you will have little touble with moe complicated poblems. Think About It σ σ What is the total chage on the "standad" ectangle when the chage density is and when D x y whee C and D ae constants? C y Because of the natue of fields, vey few poblems involve integation ove ectangles. In two dimensions, almost all integals will be in pola coodinates, and θ. In pola coodinates thee ae two pinciple ways of slicing a cicle: in concentic ings o in wedges. In Fig. 7.8 we slice the cicle in ings. A typical slice at adius and with width d is illustated. Keep in mind that d is vey small, and in fact goes to zeo in the limit, although we daw it much thicke. We then need to calculate the aea da of the ing. To do this, let us mentally take a pai of scissos, cut the ing and stetch it out staight. Since d is vey small, the stetched-out ing vey nealy foms a ectangle of length π and width d. Thus the aea of an annula ing is just (7. Aea of an annula ing) da π d. d a B d Figue 7.8 Slicing a disk into annula ings. 15

16 We can then integate to get the total aea of the cicle: A da π d π a Notice that fo any integand which depends only on, this is the appopiate way to slice the disk. Fo almost all applications using pola coodinates in this book, we will make use of this technique, so lean it well! σ Example 7.. A disk of adius a has a suface chage density C whee C is a constant. Find the total chage on the disk. a 3 q dq σ da π C dθ d πca. 3 The second way of slicing a cicle is in wedges, as shown in Fig This time the aea of the wedge is appoximately dθ that of a tiangle with base a (emembe ac length is angle times adius) and height a. Thus da ½ a dθ. Again, we can integate ove to find the total aea of the cicle. A a π 1. a This, of couse, is the appopiate way to slice a cicle wheneve the integand is only a function of θ. dθ π a. 16

17 d a Figue 7.9 Slicing a disk into wedges. Think About It When the integand depends on both and θ, show that the aea of an appopiate slice is da d dθ. Things to emembe: By appopiate slicing of aeas and volumes, you educe many integals to a simple fom. You must integate, howeve, ove all vaiables that appea in the integand. If you slice a disk into annula ings, the aea of each ing is da π d. 7.8 Thee-dimensional Integation It becomes somewhat moe complicated to conside integation in thee dimensions athe than two; howeve, the basic pinciples ae the same. In Catesian coodinates, the convesion fom two to thee dimensions is tivial. Hee the volume element is just dv dx dy dz. If we have a ectangula pism (a box shape, that is), we can find which coodinates appea in the integand and slice the pism into elements of volume appopiately. Think About It ρ What is the total chage of a cube of length 1 m on each side if the chage density is A z whee A 3 C/m 4 and z is the distance fom one specific face of the cube? 17

18 and da dv π l In cylindical coodinates, integation is vey simila to pola coodinates. Let us find the volume of a cylinde of length adius R. We put the cente of one end the cylinde on the oigin and its axis along the z axis. Now we have to divide the cylinde into elemental volumes. If the integand only depends on, the usual case, the appopiate way of doing this is by slicing the cylinde into cylindical shells. That is, we slice the cylinde with each volume element being a hollow pipe. The volume of such a cylindical shell is just the volume of a pipe (not the volume inside the pipe) of length l, adius, and wall thickness d. This is just dv l d. Notice that we have conveted a thee-dimensional integal into a one-dimensional integal. The integation is staightfowad: V R dv π l d π R l. In spheical coodinates, we will mostly deal with integands which ae functions of only. In this case, we must slice the sphee into concentic spheical shells. We may think of a typical shell of adius and thickness d as being simila to an oange peel. If we cut such a peel off an oange and spead it out flat, it is clea that the volume will be appoximately the thickness of the peel times its suface aea. Thus we have dv 4π d. We obtain the total volume of the sphee by integating fom to the adius of the sphee, R. This becomes R 4 3 dv 4π d π R. 3 V Rathe than memoizing the fomulas fo diffeent classes of integals, you should wok though a lage numbe of examples until ρ σ λ dl you have a feeling fo how to set up integals. To find total chage, we need only emembe the definition of the diffeent chage densities to get dq : dq in one dimension dq in two dimensions dq in thee dimensions. Things to emembe: The volume of a cylindical shell is π l d. The volume of a spheical shell is 4π d. Memoize these expessions, as we will make fequent use of them! 18

19 7.9 Gauss's Law and Electic Flux The electic field of any chage distibution could be detemined by summing up the contibutions of all its component chages. We essentially did this when we integated ove theads in Lesson 1. Howeve, thee ae a vey limited numbe of cases fo which closed-fom solutions ae obtainable by diect integation. Ealie in this chapte, we leaned how we can use Gauss's law in conjunction with symmety consideations to detemine the electic fields of chage distibutions with adial spheical symmety. Gauss's law can also be used to calculate the electic fields of seveal othe types of chage distibution. Peviously we used Gauss's law as a qualitative help in making ou aguments. At this point, we will constuct a mathematical fom fo Gauss's law so that we can use it moe pecisely. We know that Gauss's law states: "The net numbe of electic field lines passing though a Gaussian suface is dependent only on the chage contained within the suface." We also know that one of the fundamental chaacteistics of electic field lines is that the numbe of field lines pe unit aea is popotional to the stength of the field. We must now tun this idea into a mathematical expession. A Figue 7.1 Electic field lines passing though a fame pependicula to the field. Φ EA. Let s take a unifom electic field pointing in the x diection. We may visualize the electic field passing though a "pictue fame" which is oiented so that the electic field is pependicula to the plane of the fame. The aea of the fame is A. We know that the magnitude of the field, E, is given by the elation: E numbe of field lines, A 19

20 o, solving fo the numbe of field lines passing though the fame, numbe of field lines % EA. Since we tend to be moe comfotable woking with equalities than popotionalities, let us ewite this as numbe of field lines kea whee k is a constant that depends on the numbe of field lines we wish to associate with each Coulomb of chage. We call the quantity EA the "electic flux" and denote it with an uppe case phi, Φ. (7.3 Flux though a pependicula fame) Φ In tems of flux, EA numbe of field lines kφ. If we e not paticulaly woied about the value of the constant k, we can think of the flux as just the numbe of field lines. The units of flux ae ( V / m ) ( m ) V m. A θ Figue 7.11 Flux though a tilted fame. Φ EA cos θ. This is pefectly good if the plane of the fame is pependicula to the field lines; howeve, if the plane is tilted with espect to the field, the numbe of field lines passing though the fame is educed. To account fo this effect, it is convenient to use the definition of the aea vecto. To give a diection to an aea, we choose a unit vecto that is pependicula to the

21 the θ θ suface. (A unit vecto lying in the suface of the plane could point in any numbe of diections!) Let us denote by angle between this unit vecto and the electic field. Fo a pependicula plane,, o the field is paallel to the aea vecto. With this definition of θ, we see that the numbe of field lines passing though the fame must be popotional to cos θ. Thus: (7.4 Flux though a tilted fame) Φ EA cos θ. We can wite this esult in a shote fom by using the dot poduct. The meaning is no diffeent, howeve: Φ E A. Think About It Thee ae eally two diections which ae pependicula to the fame, so thee ae two equally logical choices fo the aea vecto. Does it make any diffeence which one we choose? This fomula is getting bette, but it is still only useful when the electic field is a constant ove the aea of the fame. If the electic field is non-unifom, we can divide the aea into small pieces, da. As long as da is small enough that E is nealy constant, we know the flux though da is d Φ E da. The total flux though the aea is then just the sum ove all such contibutions. Finally, we have a geneal equation fo flux, o the numbe of field lines passing though an aea: Electic Flux (7.5) Φ dφ E da whee Φ is the electic flux passing though a suface. It has units of volt metes (Vm). E is the electic field in volts/mete (V/m). da is an element of aea on the suface ove which we integate. The diection of the aea is nomal (pependicula) to the suface. Note that we must take the dot poduct of E and da fo each slice of aea. In geneal, this is a vey complicated integal. You should thooughly undestand what it means, but I don t expect you to have any idea how to evaluate this kind of integal in geneal. 1

22 is Φ Things to emembe: The mathematical function that descibes the numbe of field lines is called the flux. You should emembe that flux is essentially field H aea. In geneal, we need to account fo the angle of the suface with espect to the field and non-unifomities in the field when we calculate the flux. Know the fomal definition of flux Φ E da 7.1 The Integal Fom of Gauss's Law Gauss's law says that the numbe of electic field lines passing though a Gaussian suface is popotional to the chage enclosed within the suface. Now that we have a mathematical expession fo the flux and know that flux is popotional to the numbe of field lines, we can wite Gauss's law simply: b q enc Φ whee the total flux though the Gaussian suface, b is a constant, and q enc is the total chage enclosed within the suface. In tun, we can wite the flux as Φ E da whee the symbol means integation ove the entie closed (Gaussian) suface. When we have closed sufaces, we always define the diection of da to be pependicula to the suface and diected outwad fom the enclosed volume. To see how Gauss s law woks, let s apply it to a point chage, q. We know the electic field of a point chage, so we can diectly evaluate the flux. Gauss s law woks fo any Gaussian suface, but we want to make the choice of suface as easy as possible, so let s use a sphee of adius a. Because the electic field is pependicula to the sphee eveywhee, E da is just E da. We then can substitute the Coulomb's law expession fo E into the integal. We then have: Φ 1 q E da EdA 4π da a 4πε a 4πε a Φ As we noted above, Gauss's Law tells us that b q enc. Since we have a point chage, the enclosed chage is all of q. Compaing the two esults, we see that the constant b 1 / ε. 1 q q ε

23 Knowing this constant, we can wite Gauss's law in its integal fom: Gauss's Law of Electicity (7.6) Φ E E da q enc ε In this fom, Gauss's law looks quite complicated. Thee ae a few things, howeve, we should keep in mind. Fist, the concept embodied in this mathematical fom is just the same as we have aleady known: the net numbe of electic field lines passing though a Gaussian suface is popotional to the enclosed chage. Secondly, Gauss's law even in integal fom is anothe way of stating the physics of Coulomb's law. Thidly, we will only evaluate the flux integal in cases whee it is vey simple. Befoe we discuss uses of Gauss's law, we should emphasize one thing: Gauss's law is always tue, even if we do not know how to evaluate the integal. If we put a chage q inside a sphee, the flux though the suface of the sphee is q / ε. If we put a chage q inside a cube, the flux though the suface of the cube is q / ε. If we put a chage q inside a tetahedon, the flux though the suface of the tetahedon is q / ε. Theefoe, if you ae asked to find a flux though a suface, you need not evaluate a complicated integal as long as you know the total chage contained within the suface! Think About It When we discussed Gauss's law ealie, we defined the net numbe of field lines passing though a suface as the "numbe of field lines coming out of the suface minus the numbe coming in." How does the flux integal incopoate this concept? Things to emembe: Gauss s law is always tue. Gauss s law in integal fom is Φ E q E da enc ε. This is just a mathematical way of saying that the net numbe of electic field lines passing though a Gaussian suface is popotional to the chage enclosed in the suface 7.11 Applying Gauss's Law By Gauss's law, if we know the electic field eveywhee on a closed suface, we can deduce how much chage is contained within the suface. Usually, howeve, we know the 3

24 distibution of chage and we want to find the electic field. Unfotunately we can not often use Gauss's law to do that. In a few cases whee thee is a high degee of symmety; howeve, Gauss's law can be used to find electic fields. To do this, thee ae two conditions that must be met: We must choose the Gaussian suface such that the electic field is pependicula to the suface (that is, E is paallel to da ). This allows us to simplify the integand: E da E da. Note that if the Gaussian suface is pependicula to the electic field, the Gaussian suface must also be an element of a field contou. The poblem must have symmety such that the magnitude of the electic field is constant ove the entie suface (o at least the entie pat of the suface though which flux passes). With these simplifications, the flux integals becomes Φ E da E da E da EA. In othe wods, fo these special conditions and these conditions include eveything we e going to do the flux educes to what we said it was in the beginning: Electic Flux the Pactical Fom (7.7) Φ EA. This means that we can wite Gauss s law as: Gauss s Law the Pactical Fom (7.8) whee q EA enc. ε we choose a Gaussian suface that is pependicula to the field and on which the electic field is unifom (only cases of high symmety), and E is the electic field on the Gaussian suface in volts/mete (V/m). A is the aea of the Gaussian suface in squae metes (m ). ε is the pemittivity of fee space. q is the total chage enclosed in the Gaussian suface. It has units of coulombs (C). enc 4

25 One of the most impotant things fo you to lean this semeste is how to apply Gauss s law in the few cases whee thee is enough symmety that we can use Eq We will now go ove each of these cases in tun. Be sue you can epoduce these examples in detail! A. Chage Distibution with Radial Spheical Symmety A sphee of adius a has a volume chage density ρ() that has adial spheical symmety. Because of this symmety, we ecognize that the field is adial and that the pependicula (equipotential) sufaces ae concentic sphees. Futhemoe, the magnitude of the electic field must be the same at evey point on the spheical suface. We choose a sphee of adius as the Gaussian suface. It has an aea of A 4π. Thus, qenc 1 1 E q ε A 4πε 1 1 4πε 1 ε max max ρ( )4π ρ( ) It is impotant to emembe that A is the aea of the Gaussian suface and that the integal is ove all the volume included within the Gaussian suface. If we want to find the electic field inside the chage distibution whee <a, then max. If, howeve, we want to find the electic field outside the chage distibution whee >a, then max a, the adius of the chage distibution. If we e finding the field on the outside of the chage distibution, we stop integating at a because thee is no chage at lage than a. Note that this is the same conclusion we eached ealie using qualitative aguments. enc d d a a (a) Figue 7.1 Choosing a Gaussian suface to find the E field inside (a) and outside (b) a spheically symmetic chage distibution. The egion of integation is the coss-hatched geen aea. 5 (b)

26 B. Chage Distibution with Radial Cylindical Symmety Let us now find the electic field of an infinitely long cylinde of chage. The pependicula (equipotential) sufaces ae now concentic cylindes. Since it is difficult to wok with infinitely long Gaussian sufaces, let us take just as a Gaussian suface a cylindical shell of length and adius. A Gaussian suface must be a closed suface, so it must include the ends of the cylinde also. By symmety, we know that the electic field lines must adiate diectly outwad fom the axis of the cylinde. This means that no electic field lines pass though the ends of the cylinde. Because of this, only the "sides" of the cylinde contibute to the flux. Futhemoe, on the sides of the cylinde, the electic field has constant magnitude. Because of that, the flux simplifies to EA with A πl. Thus: qenc 1 1 E q ε A πε l 1 1 πε l 1 ε max max enc ρ( )π l d ρ( ) d Note that the length of the Gaussian suface, l, divides out of the poblem, so the length of ou Gaussian suface eally doesn t matte. l C. Infinite Plane of Chage One othe geomety with sufficient symmety to apply Gauss's law is an infinite sheet of chage. Let the suface chage density of the plane be σ. Hee, symmety equies the electic 6

27 (on a field to be pependicula to the plane eveywhee in space. Thee can be no component up o down, left o ight, as thee is nothing special about any given diection. The pependicula sufaces ae planes paallel to the chaged plane. We can constuct σ as a Gaussian suface anything that has two sufaces paallel to the plane; a cube of side a is a simple choice. We oient the cube so that the plane passes though the cube's cente with two faces paallel to the plane. Because the electic field is pointing outwad fom the plane, the only flux is though the sufaces paallel to the plane. Each suface has outwad (hence positive) flux of Ea, so the net flux is twice that. The chage enclosed in the cube is q enc. Gauss's law then gives: σ a σ E. ε a ε Note that this is a constant, depending only on the suface chage density. The field doesn t get weake as you move away fom the plane. This is consistent with the fact that the electic field lines don t get fathe apat as we move away fom the plane. E a D. Infinite Conducting Sheet Figue 7.7 Gaussian suface fo a plana chage distibution. σ If, instead of an infinite plane of chage, we have an infinite conducting sheet with a suface chage density of one suface of the conducto), we can also find the electic field. The only thing diffeent fom the pevious case is that in the conducting mateial thee is no electic field. Instead of placing the cube so that the faces ae outside the plane, we can place the cube with one face in the conducto and one outside. Thus the net flux out of the cube is Ea, thee being only one suface with field passing though it. The esult then is twice the pevious esult: 7

28 E σ ε The facto of esults because, in a conducting plane, each outside suface must have a suface chage density of σ. This is because the chages epel and distibute out evenly ove both sufaces. A conducting plane is then like two non-conducting paallel planes placed close to each othe. E. Some Geometical Conclusions Notice that Gauss's law tells us that the total electic flux is independent of distance fom a chage. Anothe way of stating this is that electic field lines can only be ceated o destoyed on chages. The aea of spheical Gaussian sufaces goes as, so the electic field must go as to keep the flux constant. In geneal, we can conclude that the electic field must always be invesely popotional to the aea of a Gaussian suface. The following table summaizes this fo the cases we have consideed in this chapte: Souce of Chage Dimensionality of Souce Dependence of Suface Aea Dependence of Field Point Line Plane Things to emembe: q In a few cases of high symmety, Gauss s Law educes to EA enc. ε These cases ae limited to chage distibutions with spheical, cylindical, o plana symmety. In these cases, we can use Gauss s Law to find electic fields. The only ticky pat is integating ove the chage distibution to find the enclosed chage. This is a vey impotant section. Know it well! 8

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