Chapter 14 GAUSS'S LAW

Transcription

1 Ch. 14--Gauss's Law Chapter 14 GAU' LAW A.) Flux: 1.) The wrd flux dentes a passage f smething thrugh a bundary r acrss a brder (an influx f immigrants means immigrants are passing ver a cuntry's brders). The cncept has mathematical implicatins. T understand it, an example is in rder..) Assume we have a ht, irregularly shaped bject. Assume als that there exists a vectr h that defines the directin and magnitude f the heat flw per unit area per unit time in the space arund the bject. T determine hw much heat leaves the bject per unit time, we culd use the fllwing apprach: a.) Define a clsed surface arund the bject (the surface can be any shape--it desn't have t parallel the cntur f the ht bject). b.) On the surface, define a differential area d. c.) Define a vectr nrmal (i.e., perpendicular) t the surface d ht bject differential surface d heat flw vectr h n 0 d arbitrary clsed surface arund ht bject differential surface area vectr in nrmal directin FIGURE

2 and pinted utward. Call this nrmal vectr n. d.) Define a differential surface area vectr d whse magnitude equals the differential surface area d and whse directin is perpendicular t the surface's face (i.e., in the directin f n). Mathematically, this vectr will equal: d dn. e.) Figure 14.1 n the previus page shws all f ur defined quantities: the ht bject, the arbitrary surface, the nrmal vectr n, the differential surface d, the differential surface area vectr d, and the heat flw vectr h. f.) The heat flw frm the bject per unit time will equal the heat flw thrugh the surface per unit time. This is called the heat flux Φ h (per unit time) thrugh. The heat flux (per unit time) can be deduced by deriving a general expressin fr the heat flw thrugh the differential surface d, and then by integrating that expressin ver the entire surface. g.) By the definitin f a dt prduct, h. d gives us the cmpnent f h in the directin f d (the cmpnent f h perpendicular t d prduces n flux thrugh the surface), r the amunt f heat that passes thrugh d (per unit time). The units are: (heat flw/unit area/unit time)(unit area) heat flw/unit time. h.) Integrating the dt prduct gives us the ttal heat flux (per unit time) thrugh the entire surface. Mathematically, this is written: Φ h h d. Nte: This integral des nt shw limits as usual. Instead, the letter "" is used t shw that the integral is a summatin ver a surface. In ther wrds, the actual limits will depend upn the gemetry f the surface itself (yu will see why this isn't a prblem shrtly). Nte als that the capital Greek letter phi is used t dente flux, and that that symbl is subscripted t dente what kind f flux it is. In this case, it is a heat flux. 3.) We have examined a situatin in which there is a vectr-defined quantity (heat flw) that has passed acrss a bundary, hence creating a 44

3 Ch. 14--Gauss's Law flux. What is imprtant is that any vectr quantity can have a flux assciated with it. B.) Gauss's Law in General: 1.) metime arund the turn f the 1800's, Carl Friedrich Gauss made sme interesting bservatins cncerning electric fields. pecifically, he suggested that the electric flux generated by an electric field passing thrugh a clsed surface must be related slely t the charge enclsed within the surface. His reasning fllws: a.) If an imaginary, clsed surface is defined in the vicinity f an electric field and there is n net charge inside the surface, the electric field lines entering the surface will exit at sme pint and the net electric flux thrugh the surface will be zer (see Figure 14.). b.) If, n the ther hand, there is a charge inside the surface, then an excess f electric field lines will either enter r exit the surface (depending upn whether the net charge is psitive r negative--remember, electric field lines leave psitive charges, enter negative charges) and the electric flux will nt be zer (see Figure 14.3). electric field lines passing thrugh a surface in which there is n net charge (the flux in equals the flux ut) FIGURE 14. electric field lines passing thrugh a surface in which there is a net charge (nte the net flux) +Q FIGURE 14.3 c.) As the electric flux thrugh the surface is dependent upn the charge inside the surface, the electric flux must be prprtinal t that enclsed charge. After determining the prprtinality cnstant, Gauss was able t write the relatinship as: E d q enclsed, where is the permittivity f free space (8.85x10-1 C /jule. m). 45

4 d.) The abve expressin is called Gauss's Law. It is a very pwerful tl fr deriving electric field functins when symmetry is present. C.) Gauss's Law in Practice--pherical ymmetry: 1.) Cnsider a single pint charge Q sitting in space. We will use Gauss's Law t determine an expressin fr the electric field a distance r units frm the field-prducing charge Q..) Gauss's Law is predicated n the assertin that if a charge is enclsed inside an imaginary surface, the net electric flux thrugh the surface will equal the net charge enclsed within the surface divided by a cnstant. This assertin is always true, thugh ne has t be careful hw the relatinship is used. T see this: a.) Cnsider an imaginary spherical surface (this is called a Gaussian surface) asymmetrically psitined abut a charge Q. b.) Gauss's Law des wrk in this situatin. Unfrtunately, due t the asymmetry, bth the angle between E and d and the magnitude f E vary frm pint t pint ver the surface (see Figure 14.4). In ther wrds, integrating the dt prduct between E and d des yield a numerical result that equals Q/, but the mathematics is s cumbersme that its executin is almst impssible t accmplish. spherical Gaussian surface (with electric field vectrs shwn as evaluated at different pints n the surface) d E 1 Q d E FIGURE ) A mre intelligent way t apprach this prblem is t place the Gaussian surface symmetrically abut the charge (see Figure 14.5). Ding s yields a number f useful cnsequences: a.) The electric field vectr E is the same magnitude at every pint n the surface. spherical Gaussian surface symmetrically placed d E Q E d E d FIGURE

5 Ch. 14--Gauss's Law b.) The vectrs E and d are parallel t ne anther at every pint n. 4.) With this placement f the Gaussian surface, we can easily use Gauss's Law as fllws: a.) Draw the apprpriate Gaussian surface, given the symmetry f the charge cnfiguratin (we have already dne this abve). b.) Gauss's Law states: E d q enclsed. c.) There are tw distinct peratins ging n in the evaluatin f this equatin. The right-hand side f the expressin requires ne t determine the amunt f charge enclsed within the Gaussian surface. In ur prblem, that is easy--it is simply Q. The left-hand part f the expressin requires us t evaluate the integral E. d. The evaluatin f that integral is dne in steps belw s as t allw fr cmmentary. d.) Nting that E and d are in the same directin, we can write the dt prduct assciated with ur integral as: E d E(d) csθ, where E is the magnitude f E, d is the magnitude f d, and θ is the angle between the tw vectrs. As θ 0, we can re-write the expressin as: E(d). Due t the symmetry in the prblem, the magnitude f E is the same at every pint n the surface. That means we can pull the E term ut f the integral, yielding: E d. 47

6 This is great. The integral sums the differential surface areas ver the entire surface. As such, it equals the surface area f the entire sphere. Fr a sphere, this is 4πr where r is the sphere's radius. With this, we can write: E d E(4πr ). e.) Ding the prblem as yu wuld n a test: q E d enclsed Q Ed ( ) cs0 Q E ( d) Q E( 4π r ) Q E. π 4 r This is exactly the expressin we derived fr the electric field prduced by a pint charge using Culmb's Law. Big Nte #1: When dealing with spherical symmetry, the left-hand side f Gauss's equatin will always equal E(4πr ) after evaluatin. That is, yu shuld ALWAY be able t write ut the left half f Gauss's equatin, given the symmetry is spherical. f.) (Big Nte # is s large that it has been included as a lettered part f the cmmentary). The questin arises, "What wuld have happened if the charge had been negative?" There are a number f ways t apprach this. The fllwing is the prtcl we will fllw: i.) Assume the electric field E and the differential surface area d are ALWAY utward frm the surface. That means the angle between the tw vectrs will always be zer and the dt prduct will always be E(d). 48

7 Ch. 14--Gauss's Law ii.) Always include the sign f the net charge enclsed inside the Gaussian surface when determining q enclsed. That means: 1.) If the net charge is psitive, yur slutin fr the magnitude f E will be psitive and yu will knw that yur assumed directin fr the E field was crrect (i.e., the field is utward)..) If the net charge is negative, yur slutin fr the magnitude f the electric field will be negative. As magnitudes shuld never be negative, this means yu have assumed the wrng directin fr the field. uch an bservatin requires n changes n yur part. It simply trumpets the crrect directin f the net field (i.e., it is inward, nt utward). D.) Mre Fun With pherical ymmetry: 1.) Cnsider a sphere f uter radius R that has a spherical hle f radius R 1 at its center (see Figure 14.6 t the right). The vlume charge density inside the slid part f the sphere is ρ ka, where a is the distance between the sphere's center and a pint inside the slid prtin f the shell. With that infrmatin, what is E(r) fr r < R 1, fr R 1 < r < R, and fr r > R, where r is an arbitrary distance frm the sphere's center t a pint in the regin f interest? charged sphere with hle at center R1.) Fr r < R 1 : a.) Figure 14.7 shws an imaginary Gaussian surface psitined at an arbitrary distance r frm the center f the spheres. b.) Using Gauss's Law, we get: E d q enclsed 0 E 0. c.) Des this fllw? Certainly. The net charge enclsed within the Gaussian surface R FIGURE 14.6 imaginary Gaussian surface r FIGURE

8 is zer. That means the net flux thrugh the Gaussian surface must be zer. The vectr d is nt zer, s the nly way the integral E(ds) can equal zer is if E evaluated n the surface is zer. Big Nte: It is usually abut this time that smene says, "What abut all the charge utside the Gaussian surface? Why isn't it prducing any field at r?" The answer is, "It is." It just isn't prducing any NET field at r. Hw s? Remember in the chapter n gravitatinal effects when we dealt with a bdy inside the earth? We determined that the nly mass prducing a net gravitatinal frce n such a bdy is the mass inside the sphere upn which the bdy sits. Why? Because the gravitatinal attractin between tw pint masses is an inverse distance squared functin (F α 1/r ). With that kind f functin, the frces due t all the bits f mass utside the sphere add t zer. The electric field prduced by a pint charge is als an inverse-distance-squared functin. What that means is that if there is symmetry, the vectr sum f the electric fields prduced by the charges distributed utside the Gaussian surface will add t zer. mall Nte: Anther way t lk at the prblem utlined in the Big Nte abve is t remember that Gauss's Law deals with electric flux. A cmplex, symmetric charge distributin utside a Gaussian surface will prduce n net flux thrugh the surface (whatever field lines pass int the vlume defined by the surface will sner r later pass ut f the surface). If, then, yu are willing t accept Gauss's assumptin that the net flux thrugh the surface must be related t charge inside the surface, it naturally fllws that FOR YMMETRIC ITUATION, the net electric field generating the flux thrugh the Gaussian surface must be due nly t the charge enclsed by the surface. In shrt, FOR A YMMETRIC FIELD, the nly charge we ever need t wrry abut is that enclsed by the Gaussian surface. 3.) Fr R 1 < r < R : a.) Figure 14.8 shws an imaginary Gaussian surface psitined an arbitrary distance r units frm the sphere's center. b.) Befre we can use Gauss's Law, we need t determine the charge inside the Gaussian surface. T d s: imaginary Gaussian surface r FIGURE

9 Ch. 14--Gauss's Law i.) Define a differential, spherical shell f radius b and thickness db (see Figure 14.9). The differential vlume f that shell is: dv (surface area)(thickness) (4πb )db. ii.) The differential charge dq in that vlume equals the vlume charge density evaluated at b times the differential vlume, r: differential charge dq in differential vlume dv dq dv (kb)(4πb )db b dv (4πb )db db FIGURE 14.9 dq ρdv (kb)[(4πb )db]. c.) Gauss's Law yields: q E d Ed ( ) E enclsed ( d) E( 4π r ) 1 4πk E π 4 r k r dq b R1 4 b 4 () ρ dv r ( kb) [( 4πb ) db] b R1 r 3 ( b ) db b R1 4 4 k r R 1. r 4 4 r 51

10 4.) Fr r > R : a.) Figure shws a Gaussian surface psitined an arbitrary distance r units frm the sphere's center. situatin fr r > R r Gaussian surface b.) The left side f Gauss's Law is as usual fr spherical symmetry, given that r is the radius f the Gaussian surface. FIGURE c.) The right side f Gauss's Law lks very similar t the previus prblem (i.e., E(r) fr R 1 < r < R ). The nly difference is that q enclsed is all the charge in the spherical shell. i.) Cnsequence: Gauss's Law will lk exactly as it did in the previus prblem with the exceptin that the limits f integratin t determine q encl will be frm R 1 t R instead f frm R 1 t r. d.) As such, Gauss's Law fr this part lks like: q E d Ed ( ) E enclsed E( 4π r ) k r dq R b R1 4 b 4 b R1 4 4 k R R1. r 4 4 ( kb) [( 4πb ) db] R 5

11 Ch. 14--Gauss's Law E.) One Mre Twist--urface Charge Density: 1.) There is ne ther kind f charge density functin available that hasn't yet been discussed. A surface charge density functin σ defines the amunt f charge per unit area a structure has n its surface. a.) Re-cnsider the thick, spherical shell prblem dne in Part d. Alng with the vlume charge density ρ sht thrugh the slid between R 1 and R, assume an additinal surface charge density -σ has been placed n the sphere's uter surface (we are making it negative just t be different--the negative sign has been unembedded t make it clear what kind f charge is present). In that case, determine E(r) fr r < R 1, fr R 1 < r < R, and fr r > R. b.) There is t be n difference in analyzing E(r) fr r < R 1 and fr R 1 < r < R. Why? Because the extra charge has been placed utside the defined Gaussian surfaces fr bth f these regins. c.) As the prblem is nw set up, nly the field fr r > R will be different. pecifically: q enclsed ρdv - σa, where A is the area f the sphere's utside surface (this will be 4πR ). d.) Frm lking at the q enclsed expressin abve, tw things shuld be nted: i.) Always add all the charge--ign INCLUDED--inside a Gaussian surface. ii.) Yu will never encunter a surface charge density functin that is nt cnstant. Why? Because a variable density functin lses the charge symmetry required fr Gauss's Law t wrk. The cnsequence f this surface-charge-density-functinsmust-be-cnstant bservatin is that we will never have t d anything with differential area da's. 53

12 e.) Just t be cmplete, Gauss's Law fr this prblem (assuming r > R ) reads: [ ] σ(4πr [ ] E d ρdv ). F.) Cnductrs: 1.) The difference between a cnductr and an insulatr has t d with the kind f bnding utilized. a.) Insulatrs use cvalent bnding. uch bnding wrks by the sharing f uter shell (valence) electrns between neighbring atms. In insulatrs, an atm's electrns never stray farther than a few atmic radii frm their parent atm. b.) Cnductrs use metallic bnding. uch bnding wrks by the sharing f uter-shell electrns between all ther atms in the structure. In cnductrs, valence electrns can migrate freely thrughut the structure..) Free charge is accelerated when in an electric field. If there is a net, static, electric field permeating a cnductr, the valence electrns within the cnductr will mve until they psitin themselves s as t eliminate the field (i.e., they can't stp until the net field is zer). In shrt, FOR A TATIC CHARGE ITUATION, THE ELECTRIC FIELD INIDE A CONDUCTOR WILL ALWAY BE ZERO. 3.) With that in mind, cnsider the fllwing situatin: Figure shws a pint charge -Q placed at the center f a thick, hllw, cnducting spherical shell f inner radius R 1 and uter radius R (again, we are using a negative charge t be different). The cnductr is electrically neutral (that is, there is n extra charge placed n the sphere--there are as many electrns as prtns in the structure). What is the E(r) fr r < R 1 and fr R 1 < r < R? what appears t be the case a.) Fr r < R 1 : The charge enclsed is equal t -Q. Using Gauss's Law fr spherical symmetry in very brief frm yields: -Q FIGURE

13 Ch. 14--Gauss's Law q E d E enclsed Q where the negative sign means the field is inward 4π r,. b.) Fr R 1 < r < R : The temptatin is t assume the same slutin hlds fr this regin as held fr r < R 1. After all, a Gaussian surface extending int the regin between R 1 and R appears t have -Q's wrth f charge inside it (again, see Figure 14.11). If that is the case, the net charge inside the Gaussian surface shuld prduce an electric field in the regin. The prblem? The sphere is a CONDUCTOR. There can't be an electric field in the regin. Hw d we reslve this seeming paradx? c.) When the -Q charge is first intrduced at the sphere's center, it briefly generates a net electric field inside the metallic sphere. Electrns in the sphere respnd t the field by re-distributing themselves s as t eliminate the field (this takes nly a few millinths f a secnd t d). Hw did they re-distribute? -Q's wrth f valence electrns within the sphere migrate away frm the -Q at the sphere's center, leaving +Q's wrth f charge n the sphere's inside surface. In ding s, the amunt f charge inside the Gaussian surface between R 1 and R becmes zer. Where des the -Q's wrth f valence charges g? They mve t the utside surface f the sphere (see Figure 14.1). What is really the case? Electrns in the cnductr are repulsed t the utside surface by the presence f -Q at the center. That leaves +Q n the sphere's inner surface. Gaussian surface -Q -Q n utside surface d.) Bttm Line: AGAIN, THERE CAN BE NO TATIC ELECTRIC FIELD INIDE A CONDUCTOR. If a field is mmentarily induced, the charges within the cnductr will rearrange themselves s as t eliminate it. As such, a Gaussian surface lcated inside +Q n inside surface Nte: The net charge inside the Gaussian surface between R 1 and R is ZERO. FIGURE

14 a cnducting vlume will ALWAY have NO NET CHARGE within it. G.) Gauss's Law in Practice--Cylindrical ymmetry: 1.) Cnsider a very lng wire with a cnstant, linear charge density λ n it (a linear charge density functin defines the amunt f charge per unit length n a structure). As lng as ur attentin is fcused in clse arund the central sectin f the wire, we can ignre edge effects and assume the electric field is radially utward (r we culd d what mst bks d and make the highly unrealistic assumptin that the wire is infinitely lng). Use Gauss's Law t derive an expressin fr E(r), where r is an arbitrary distance radially ut frm the wire (i.e., perpendicularly ut). wire Gaussian surfaces arund line f charge r end surface 1 linear charge density cylindrical Gaussian surface end surface.) Again, assuming we are nt near the ends f the wire, the symmetry here is cylindrical. Hw s? Examine Figure It shws an imaginary, cylindrical Gaussian surface psitined at an arbitrary distance r frm the wire. d E FIGURE ) This Gaussian surface is actually three surfaces in ne--tw flat, circular surfaces at each end and the cylinder itself. E d 1 a.) Figure defines E and d fr bth the cylindrical Flux thru ends is zer; flux thru cylinder is nn-zer FIGURE

15 Ch. 14--Gauss's Law part f the structure as well as fr ne f the flat end-pieces. b.) As E and d 1 n the end are at a right angle t each ther, their dt prduct equals zer. The same is true at the ther end. Nte 1: If we were nt dealing with the central sectin f the wire, the electric field wuld nt be radially utward, there wuld be flux thrugh the end-pieces, and we wuld nt be able t use Gauss's Law t determine the electric field functin we want. Als, as the wire is nt infinitely lng, the electric field expressin we derive here will nly be gd near the wire (versus being gd far frm the wire... like at infinity). parallel cmpnents add t zer; electric field is radial in central area lng line f charge Nte : till dn't believe the field will be radially utward? Cnsider the fllwing: Fr any pint near the center f the wire, there will be electric fields generated by bits f charge all alng the wire. Take tw mirrr charges (i.e., differential charges n ppsite sides and apprximately the same distance frm the central sectin). The cmpnents f the electric field parallel t the wire will add t zer leaving nly the cmpnents radially ut frm the wire (see Figure 14.15). Nte 3: Des that mean the electric field is caused by charge carriers all alng the wire? Yes! Des that mean that the charge utside the Gaussian surface has t be taken int accunt when using Gauss's Law? NO! NO! NO! Gauss's Law desn't say that the electric field is generated by the charge inside a Gaussian surface, it says that the FLUX THROUGH THE GAUIAN URFACE is related t the charge enclsed (the net flux frm charges utside the surface must be zer). We are able t deduce the electric field assciated with that flux, whatever the accumulated charge has made that field t be, nly because there is a mathematical relatinship between an electric flux and its electric field. FIGURE ) As the electric flux thrugh the ends is zer, the nly flux present is thrugh the cylindrical part f the structure. With that in mind, we can write Gauss's Law as: E d q enc. cyl.urf 57

16 5.) Examining the integral: a.) As the angle between E and d is zer n the cylinder, the dt prduct inside the integral is simply E(d). b.) Due t symmetry, the magnitude f E is the same at every pint n the cylinder's surface. That means we can pull the E term utside the integral just as we did when we dealt with spherical symmetry. c.) The sum f the differential surface areas ver the entire cylindrical surface (i.e., d) equals the surface area f the cylinder. The surface area f a cylindrical shell is: (circumference)(length) (πr)l, where r is the radius and L is the length f the Gaussian surface. d.) Bttm Line: Fr cylindrical symmetry, the left-hand side f Gauss's Law will ALWAY equal: E(πrL). 6.) The right-hand side f Gauss's Law requires a determinatin f the amunt f charge inside the Gaussian surface. a.) In this case, the charge per unit length is λ. b.) The wire inside the Gaussian surface has a length L, s the charge inside the Gaussian surface must be: q encl (λ C/m)(L meters) λl. 7.) Writing this ut frmally (i.e., in the way yu might d n a test), we get: 58

17 Ch. 14--Gauss's Law q E d enc Ed ( ) cs0 E d E( π rl) λ E. π r λl λl λl Nte: The L terms canceled. This is gd! It wuld have made n sense t have cncluded that ur electric field was dependent upn the length f the imaginary Gaussian surface used t generate the electric field expressin. H.) Deeper and Deeper int Cylindrical ymmetry: 1.) Cnsider a lng rd f radius r 1 with a vlume charge density f kr sht thrugh it. Caxially, a cylindrical, cnducting shell f inside radius r and utside radius r 3 is placed arund the rd (see Figure 14.16). On the shell is a cnstant surface charge density -σ. Determine E(r) fr: a.) r < r 1 ; b.) r 1 < r < r ; Crss-sectin f Caxial ystem --Cylindrical ymmetry-- r 1 r r 3 cylindrical shell with surface charge density rd with vlume charge density kr FIGURE c.) r < r < r 3 ; d.) r 3 < r. 59

18 Nte: This is typically the way test prblems are stated..) Fr r < r 1 : a.) The left-hand side f Gauss's Law is always the same fr cylindrical symmetry. That is, the integral ends up equaling E(πrL). Again, we are assuming that bth E and d are utward, understanding that if this is nt true the sign f the charge will crrect the versight. b.) The right-hand side requires the use f a vlume charge density functin t determine the charge enclsed within the cylindrical Gaussian surface. Let a be the distance frm the central axis t a differentially thin, cylindrical shell f length L (i.e., the length f the Gaussian surface) and differential thickness da (t see all f this, Figure shws a blwn-up sketch f the rd by itself). The vlume is: dv (surface area f the cylinder)(thickness) (πal)da. Crss-ectin f Rd dv (surface area)(thickness) [(circumference)(length)][thickness] (πa)(l)da da a r r 1 cylindrical Gaussian surface f length L dq dv (ka)[(πal)da] FIGURE

19 Ch. 14--Gauss's Law while the differential charge enclsed in that differential vlume is: dq ρdv (ka)[(πal)da]. c.) Using this infrmatin, and nting that the charge within the Gaussian surface is distributed between a 0 and a r, we can write: 3.) Fr r 1 < r < r : q E d E Ed ( ) E ( d) E π rl) π kl ( 3 k a r 3 kr 3. encl r a 0 ρdv ( ka)[( πal) da] r a 0 a da a.) In the regin between r 1 and r, the charge enclsed inside an imaginary, cylindrical Gaussian surface f radius r will be the ttal charge inside the rd. Assuming the radius f the Gaussian surface is r, the nly difference in this prblem and the prblem in Part will be the limits f integratin. i.) Hw s? In Part we integrated frm a 0 t a r. Why? Because we needed the charge inside the Gaussian surface which was, in turn, inside the rd. As such, ur summing had t terminate inside the rd. ii.) In this case, ur Gaussian surface is utside the rd. That means we must terminate ur integratin where the charge ceases t exist (i.e., at r 1 ), nt at r. 61

20 b.) In a smewhat truncated frm, ur derivatin fr this sectin fllws as: E d () ρ dv E( π rl) Er () k r r1 a 0 r1 a 0 3 k r 1. r 3 ( ka)( πal) da ada 4.) Fr r < r < r 3 : a.) The shrt answer: The field must be zer as the Gaussian cylinder resides inside a cnductr. b.) The lng answer: Psitive charge inside the rd attracts electrns inside the cnductr t the cnductr's inner wall. Hw much charge? An amunt equal t the amunt f psitive charge n the rd. Cnsequence: The net charge inside the Gaussian surface is zer and the net field evaluated n the surface f the Gaussian cylinder must als be zer. 5.) Fr r 3 < r: Crss-sectin f cnfiguratin and Gaussian surface fr field utside the caxial system a.) Figure shws the system and the Gaussian surface that is apprpriate t this prblem. b.) The charge enclsed in the Gaussian surface cmes frm the psitive charge distributed inside the rd (i.e., frm a r Gaussian surface L FIGURE

21 Ch. 14--Gauss's Law 0 t a r 1 ) and the net negative charge that was initially placed n the cnductr (the induced charge n the inside and utside surfaces f the cnductr will add t zer). c.) Using Gauss's Law, we can write: q E d encl E( π rl) r1 a ( ka)( π al) da + ( σ) A 0. d.) The integral prtin f q enclsed is the same as in Part 3. e.) A is defined as the area f that prtin f the utside surface f the cylindrical shell fund inside the Gaussian surface. Mathematically, it is A (circumference)(length) (πr 3 )L. f.) Ding the integral, substituting in fr A, and dividing ut the πrl terms leaves: E k r 3 1 σr3 3. r I.) Insulating heets: 1.) A thick, rectangular, insulating slid is permeated with a unifrm charge density thrughut its vlume (see Figure 14.19). a.) We culd assign a vlume charge density t the bject and g frm there. b.) An alternative apprach wuld be t define a surface area charge density functin (i.e., σ) fr the structure. That is, we culd assciate the amunt f charge in a given vlume with the surface area charge-filled slab charge-filled vlume vlume's face (area A) (charge in vlume)/a FIGURE

22 f the vlume's face (again, see Figure 14.19). Mathematically, this wuld be: σ (charge inside vlume)/(face area f vlume). c.) The ratinale fr creating such a functin is simple. With it we can use Gauss's Law t derive an expressin fr the electric field just utside any unifrm, charge-filled insulatr. Hw s? Cntinue n!.) Cnsider a very thin, flat sheet f charge with a surface charge density σ culmbs per square meter assciated with it. What is the electric field clse t the surface and arund the central sectin f the sheet? 3.) Figure 14.0a shws the set-up alng with a cubic Gaussian surface, where A is the face area f the surface. 3-d view side view sheet f charge E A A E Gaussian surface Gaussian surface face area A sheet f charge FIGURE 14.0a FIGURE 14.0b Nte: As lng as the sides f the Gaussian surface are either parallel r perpendicular t the sheet itself, the shape f the Gaussian surface is incnsequential. That is, we are using a cube here; it culd as well have been a cylindrical plug. 4.) Ignring edge effects (we are dealing with the area in the central sectin f the sheet) and assuming we are lking fr the field very near the 64

23 Ch. 14--Gauss's Law sheet, the electric field will be perpendicularly ut frm the sheet. That means: a.) The flux thrugh each side-face (that is, each face perpendicular t the plane f the sheet) is zer. b.) Assuming the cube is psitined s that the parallel faces are symmetrically placed abut the sheet: i.) The electric field passing thrugh bth faces will have the same magnitude (nte the directin thrugh bth faces is utward). ii.) Examining the side-view f the system in Figure 14.0b, the dt prduct used t determine the flux thrugh each f the tw parallel faces will be psitive and equal t EA. 5.) Gauss's Law yields: q E d encl EA + EA σ E. σa 6.) This expressin is always true whenever we want the electric field very near the central sectin f a sheet f charge r a charge-filled insulating slab fr which we have defined a surface charge density functin. 7.) Des it make sense fr this functin t be independent f the distance frm the surface? Yes! Hw s? Fllw alng: a.) When the pint f interest (call it Pint A) is very near: i.) Charges n the surface clse t Pint A will prduce large electric fields perpendicularly ut frm the surface. They will als prduce small electric fields parallel t the surface, but thse 65

24 will add t zer when all the near-charge fields are vectrially added. ii.) Charges n the periphery will prduce smaller net electric fields. Thse fields will have large cmpnents parallel t the surface which will add t zer when vectrially summed. They will als have smaller cmpnents perpendicularly ut frm the surface. Figure 14.1 shws all. Pint A very near charge surface a peripheral charge near-charge net nearcharge field sheet f charge peripheral-charge fields a peripheral charge iii.) When Pint A mves ut away frm the surface, the amunt FIGURE 14.1 f field prduced by the near-charge fields will diminish. The peripheral-charge field will als diminish, but the perpendicular cmpnent f thse charges will becme prprtinally larger (see Figure 14.). The net effect is that the net electric field desn't change. iv.) This wrks nly when relatively clse t the surface. Hw clse? Gd questin. If yu have nthing better t d, try t determine hw far ut frm a 1 meter radius, circular surface (this gemetry will be easier t deal with) yu can get and still have the actual electric field value be within 5% f σ/. mving a shrt distance away frm sheet (net field remains the same) a peripheral charge peripheral field near-charge near-charge field diminishes perpendicular cmpnent f peripheral-charge field gets prprtinally larger parallel cmpnents f peripheral-charge field still add t zer a peripheral charge FIGURE

25 Ch. 14--Gauss's Law Nte: If this were really imprtant, I'd have an answer. It isn't, s instead f an answer yu get a prblem at the back f the chapter. J.) Cnducting heets: 1.) A thick, rectangular, cnducting slid has charge unifrmly distributed ver its entire surface (assume the surface charge density is σ). What is the electric field clse t the sheet and arund its central regin? 3-d view side view face area (inside the cnductr) charge-cvered cnductr E 0 inside cnductr A Gaussian surface E Gaussian surface surface charge density (cvers whle bdy) face area A (just utside the charged surface) n flux thrugh this surface as E 0 cut-away A FIGURE 14.3a FIGURE 14.3b.) There are tw ways t d this. The first is embdied in Figures 14.3a and 14.3b. A Gaussian plug (as befre, it happens t be a cube, thugh it culd have been a cylindrical plug) is placed s that ne face is inside the cnductr. a.) As the electric field inside a cnductr is zer, the flux thrugh the left face will be zer. That leaves flux passing nly thrugh the right face. b.) Using Gauss's Law, we get: 67

26 q E d encl EA E σa σ. c.) Ntice that this expressin is very similar t the ne derived fr a thin sheet f charge (r an insulating slab with a surface charge density functin assciated with it). 3.) An alternative apprach wuld be t extend the Gaussian plug s that the end faces are the same distance frm their respective uter surfaces (see Figure 14.4). a.) In that case, there will be a flux thrugh bth end faces, but there will als be twice as much charge within the Gaussian surface (there is σ wrth f charge per unit area n all uter surfaces f the structure). b.) As such, Gauss's Law becmes: side view E A E 0 inside cnductr cut-away A Gaussian surface E FIGURE 14.4 q E d encl EA + EA σ E. σa+ σa 68

27 Ch. 14--Gauss's Law QUETION 14.1) A ski cap shwn t the right in Figure I has an pen hle whse area is A. The ttal surface area f the cap is 5A. If the cap's hle is psitined s that it is in the x-z plane and an electric field in the y-directin permeates the regin in which the cap resides, what is the net electric flux thrugh the cap's surface? z y E x FIGURE I ski cap 14.) A spherical insulatr f radius R has Q's wrth f charge unifrmly distributed thrughut it. Determine the electric field E(r) fr: a.) r < R, and; b.) r > R. 14.3) A spherical, cnducting shell has an inside radius R 1 and utside radius R. Q's wrth f charge is smehw levitated at the shell's center while a charged rd is used t initially place Q's wrth f charge n the shell's inner surface: a.) After a shrt time, hw much charge will end up n the inner surface f the shell? b.) Hw much charge will end up n the uter surface f the shell? c.) Derive an expressin fr E(r), where r < R 1 ; d.) Derive an expressin fr E(r), where R > r > R 1 ; and e.) Derive an expressin fr E(r), where r > R. f.) Out f curisity, what is the surface charge density n the uter surface f the shell? 14.4) A spherical shell has an inside radius R 1, an utside radius R, and a vlume charge density defined by (C/r )e kr (nte that r is the distance frm the shell's center t a vlume f interest, k is a cnstant equal t ne and having the apprpriate units, and C is als a cnstant having the apprpriate units). Layered ver this shell is a cnducting shell whse 69

28 inside radius is R, whse utside radius is R 3, and upn which has been placed a cnstant surface charge density σ. a.) What are k's units; what are C's units? b.) Accurately plt the Electric Field versus Psitin graph fr this charge cnfiguratin. Nte that this is nt a gee whiz questin. Use Gauss's Law t determine the electric field expressin fr each regin, then graph it. Fr the sake f simplicity, assume that C 10-1, σ 10-1 culmbs per square meter, R 1 1 meter, R meters, and R 3 3 meters. Use these values nly after yu have derived yur general expressins. Nte: DO NOT TAKE A LOT OF TIME ON THE GRAPH. WHAT I IMPORTANT HERE I UING GAU' LAW TO DERIVE THE GENERAL ELECTRIC FIELD EXPREION FOR EACH REGION. c.) Lking at the graph, wuld yu say that electric field functins are cntinuus? 14.5) A thick-skinned pipe f inside radius r 1 and utside radius r has a surface charge density σ 1 n its inside surface and a surface charge density σ n its utside surface (σ 1 σ, r 1 and the charge assciated with σ is negative). A secnd thick-skinned pipe surrunds the first pipe caxially. Its inside radius is r 3, its utside radius is r 4, and it has a vlume charge density ρ r 3 r 4 r 1 FIGURE II Ce kr r (C/r)e kr sht thrugh its vlume. Figure II shws the layut. a.) Given the infrmatin prvided, can yu tell fr sure whether either pipe is a cnductr r an insulatr? b.) Accurately plt the Electric Field versus Psitin graph fr this charge cnfiguratin. Fr the sake f simplicity, assume that C 10-1 culmbs per square meter, σ culmbs per square meter, σ -x10-1 culmbs per square meter, r 1 1 meter, r meters, r 3 3 meters, and r 4 4 meters. Use these values nly after yu have derived yur expressins. Nte: As in Prblem 14.4, d nt take a lt f time n the graph. 70

29 Ch. 14--Gauss's Law 14.6) A very lng cylindrical rd f radius R has an unknwn vlume charge density sht thrugh it. The electric field a distance r units frm the rd's central axis is fund t be E kr 6 /(6 r), where k is a cnstant and r > R. Use Gauss's Law and yur head t determine TWO different vlume charge density functins that culd prduce this field. R r FIGURE III 14.7) A thick, insulating disk f radius R 1 meter and thickness t.0 meters has Q's wrth f charge unifrmly distributed thrughut its vlume. a.) Use Gauss's Law and the apprpriate vlume charge density functin t derive an expressin fr the electric field very clse t the disk's surface and alng the disk's central axis. b.) Determine an equivalent surface charge density functin fr this disk. c.) (This is mre f a review questin than anything else.) Using the surface charge density functin, treat the disk like a sheet f charge. Use Gauss's Law t determine the electric field dwn the central axis and very clse t the disk's surface. As this field is accurate nly when near the disk, call it the near-pint field. d.) Using the ld-fashined definitin apprach t determine the electric field dwn the disk's axis (i.e., define a differential pint charge n the sheet f charge, determine the field due t that pint charge, then integrate t determine the net field due t all the charge). Call this the general, axial field. e.) The near-pint field is an apprximatin that is gd nly when near the disk. The general, axial field is gd at any pint alng the axis. Hw far frm the disk can ne get and still have the electric field value determined using the near-pint field expressin be within 5% f the electric field value determined using the general, axial field expressin? 71

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