Chpte The lectic Field II: Continuous Chge Distibutions Conceptul Poblems [SSM] Figue -7 shows n L-shped object tht hs sides which e equl in length. Positive chge is distibuted unifomly long the length of the object. Wht is the diection of the electic field long the dshed 45 o line? xplin you nswe. Detemine the Concept The esultnt field is diected long the dshed line; pointing wy fom the intesection of the two sides of the L-shped object. This cn be seen by dividing ech leg of the object into (o moe) equl segments nd then dwing the electic field on the dshed line due to the chges on ech pi of segments tht e equidistnt fom the intesection of the legs. Positive chge is distibuted unifomly long the entie length of the x xis, nd negtive chge is distibuted unifomly long the entie length of the y xis. The chge pe unit length on the two xes is identicl, except fo the sign. Detemine the diection of the electic field t points on the lines defined by y x nd y x. xplin you nswe. Detemine the Concept The electic fields long the lines defined by y x nd y x e the supeposition of the electic fields due to the chge distibutions long the xes. The diection of the electic field is the diection of the foce cting on test chge t the point(s) of inteest. Typicl points e shown t two points on ech of the two lines. y y x y x + + + + + + + + x 89
9 Chpte Tue o flse: () (b) (c) The electic field due to hollow unifomly chged thin spheicl shell is zeo t ll points the shell. In electosttic equilibium, the electic field eveywhee the mteil of conducto must be zeo. If the net chge on conducto is zeo, the chge density must be zeo t evey point on the sufce of the conducto. () Tue (ssuming thee e no chges the shell). (b) Tue. The chges eside on the sufce of conducto. (c) Flse. Conside spheicl conducting shell. Such sufce will hve equl chges on its inne nd oute sufces but, becuse thei es diffe, so will thei chge densities. 4 If the electic flux though closed sufce is zeo, must the electic field be zeo eveywhee on tht sufce? If not, give specific exmple. Fom the given infomtion cn the net chge the sufce be detemined? If so, wht is it? Detemine the Concept No, this is not necessily tue. The only conclusion tht we cn dw is tht thee is equl positive nd negtive flux. Fo exmple, the net flux though Gussin sufce completely enclosing dipole is zeo. If the electic flux is zeo though the closed sufce, we cn conclude tht the net chge the sufce is zeo. 5 Tue o flse: () (b) Guss s lw holds only fo symmetic chge distibutions. The esult tht eveywhee the mteil of conducto unde electosttic conditions cn be deived fom Guss s lw. () Flse. Guss s lw sttes tht the net flux though ny sufce is given byφ da 4πk. While it is tue tht Guss s lw is esiest to pply to net S n symmetic chge distibutions, it holds fo ny sufce. (b) Tue. Becuse the chges on conducto, unde electosttic conditions, eside on the sufce of the conducto, the net flux the conducto is zeo. Hence, by Guss s lw, the electic field the conducto must lso be zeo.
The lectic Field II: Continuous Chge Distibutions 9 6 A single point chge q is locted t the cente of both n imginy cube nd n imginy sphee. How does the electic flux though the sufce of the cube compe to tht though the sufce of the sphee? xplin you nswe. Detemine the Concept Becuse the net flux is popotionl to the net chge enclosed, nd this is the sme fo both sufces, the electic flux though the sufce of the cube is the sme s the electic flux though the sufce of the sphee. 7 [SSM] An electic dipole is completely closed imginy sufce nd thee e no othe chges. Tue o Flse: () The electic field is zeo eveywhee on the sufce. (b) The electic field is noml to the sufce eveywhee on the sufce. (c) The electic flux though the sufce is zeo. (d) The electic flux though the sufce could be positive o negtive. (e) The electic flux though potion of the sufce might not be zeo. () Flse. Ne the positive end of the dipole, the electic field, in ccodnce with Coulomb s lw, will be diected outwd nd will be nonzeo. Ne the negtive end of the dipole, the electic field, in ccodnce with Coulomb s lw, will be diected inwd nd will be nonzeo. (b) Flse. The electic field is pependicul to the Gussin sufce only t the intesections of the sufce with line defined by the xis of the dipole. (c) Tue. Becuse the net chge enclosed by the Gussin sufce is zeo, the net flux, given by φ da 4πk, though this sufce must be zeo. net S n (d) Flse. The flux though the closed sufce is zeo. (e) Tue. All Guss s lw tells us is tht, becuse the net chge the sufce is zeo, the net flux though the sufce must be zeo. 8 xplin why the electic field stength inceses linely with, the thn deceses invesely with, between the cente nd the sufce of unifomly chged solid sphee. Detemine the Concept We cn show tht the chge unifomly chged solid sphee of dius is popotionl to nd tht the e of sphee is popotionl to. Using Guss s lw, it follows tht the electic field must be popotionl to /.
9 Chpte Use Guss s lw to expess the electic field spheicl chge distibution of constnt volume chge density: 4πk A whee A 4π. xpess nd : s function of ρ 4 ρ V πρ Substitute fo to obtin: 4 4πk πρ 4kπρ 4π This esult shows tht the electic field inceses linely s you move out fom the cente of spheicl chge distibution. 9 [SSM] Suppose tht the totl chge on the conducting spheicl shell in Figue -8 is zeo. The negtive point chge t the cente hs mgnitude given by. Wht is the diection of the electic field in the following egions? () < R, (b) R > > R, (c) nd > R. xplin you nswe. Detemine the Concept We cn pply Guss s lw to detemine the electic field fo < R, R > > R, nd > R. We lso know tht the diection of n electic field t ny point is detemined by the diection of the electic foce cting on positively chged object locted t tht point. () Fom the ppliction of Guss s lw we know tht the electic field in this egion is not zeo. A positively chged object plced in the egion fo which < R will expeience n ttctive foce fom the chge locted t the cente of the shell. Hence the diection of the electic field is dilly inwd. (b) Becuse the totl chge on the conducting sphee is zeo, the chge on its inne sufce must be positive (the positive chges in the conducting sphee e dwn thee by the negtive chge t the cente of the shell) nd the chge on its oute sufce must be negtive. Hence the electic field in the egion R > > R is dilly outwd. (c) Becuse the chge on the oute sufce of the conducting shell is negtive, the electic field in the egion > R is dilly inwd. The conducting shell in Figue -8 is gounded, nd the negtive point chge t the cente hs mgnitude given by. Which of the following sttements is coect?
The lectic Field II: Continuous Chge Distibutions 9 () (b) (c) (d) The chge on the inne sufce of the shell is + nd the chge on the oute sufce is. The chge on the inne sufce of the shell is + nd the chge on the oute sufce is zeo. The chge on both sufces of the shell is +. The chge on both sufces of the shell is zeo. Detemine the Concept We cn decide wht will hppen when the conducting shell is gounded by thinking bout the distibution of chge on the shell befoe it is gounded nd the effect on this distibution of gounding the shell. The negtive point chge t the cente of the conducting shell induces positive chge on the inne sufce of the shell nd negtive chge on the oute sufce. Gounding the shell ttcts positive chge fom gound; esulting in the oute sufce becoming electiclly neutl. (b) is coect. The conducting shell in Figue -8 is gounded, nd the negtive point chge t the cente hs mgnitude given by. Wht is the diection of the electic field in the following egions? () < R, (b) R > > R, (c) nd > R. xplin you nswes. Detemine the Concept We cn pply Guss s lw to detemine the electic field fo < R, R > > R, nd > R. We lso know tht the diection of n electic field t ny point is detemined by the diection of the electic foce cting on positively chged object locted t tht point. () Fom the ppliction of Guss s lw we know tht the electic field in this egion is not zeo. A positively chged object plced in the egion fo which < R will expeience n ttctive foce fom the chge locted t the cente of the shell. Hence the diection of the electic field is dilly inwd. (b) Becuse the conducting shell is gounded, its inne sufce is positively chged nd its oute sufce will hve zeo net chge. Hence the electic field in the egion R > > R is dilly outwd. (c) Becuse the conducting shell is gounded, the net chge on the oute sufce of the conducting shell is zeo, nd the electic field in the egion > R is zeo. stimtion nd Appoximtion In the chpte, the expession fo the electic field due to unifomly chged disk (on its xis), ws deived. At ny loction on the xis, the field
94 Chpte R mgnitude is π k +. At lge distnces ( z >> R), it ws z shown tht this eqution ppoches k z. Vey ne the disk ( z << R), the field stength is ppoximtely tht of n infinite plne of chge o πk. Suppose you hve disk of dius.5 cm tht hs unifom sufce chge density of.6 μc/m. Use both the exct nd ppopite expession fom those given bove to find the electic-field stength on the xis t distnces of (). cm, (b).4 cm, nd (c) 5. m. Compe the two vlues in ech cse nd comment on the how well the ppoximtions wok in thei egion of vlidity. Pictue the Poblem Fo z << R, we cn model the disk s n infinite plne. Fo z >> R, we cn ppoximte the ing chge by point chge. () vlute the exct expession fo z. cm: π 9 ( 8.988 N m /C )(.6 C/m ) z μ. cm.5 5 N/C. 5 N/C + (.5 cm) (. cm) Fo z << R, the electic field stength ne n infinite plne of chge is given by: πk vlute the ppoximte expession fo z. cm: ppox 9 ( 8.988 N m /C )(.6 C/m ) π μ. 5 N/C. 5 N/C The ppoximte vlue gees to within.4% with the exct vlue nd is lge thn the exct vlue.
The lectic Field II: Continuous Chge Distibutions 95 (b) vlute the exct expession fo z.4 cm: π 9 ( 8.988 N m /C )(.6 C/m ) z μ.4 cm. 5 N/C. 5 N/C + (.5 cm) (.4 cm) The ppoximte vlue gees to within.% with the exct vlue nd is smlle thn the exct vlue. (c) vlute the exct expession fo z 5. m: π 9 ( 8.988 N m /C )(.6 C/m ) z μ 5. m.54 N/C.5 N/C + (.5 cm) ( 5. m) Becuse z >> R, we cn use Coulomb s lw fo the electic field due to point chge to obtin: k z kπ z ( z) vlute ( 5. m) : 9 π ( ) ( 8.988 N m /C )(.5cm) (.6μC/m ).m 5 ppox.5 N/C ( 5.m).54 N/C The ppoximte vlue gees, to fou significnt figues, with the exct vlue. Clculting Fom Coulomb s Lw [SSM] A unifom line chge tht hs line chge density l equl to.5 nc/m is on the x xis between x nd x 5. m. () Wht is its totl chge? Find the electic field on the x xis t (b) x 6. m, (c) x 9. m, nd (d) x 5 m.
96 Chpte (e) stimte the electic field t x 5 m, using the ppoximtion tht the chge is point chge on the x xis t x.5 m, nd compe you esult with the esult clculted in Pt (d). (To do this you will need to ssume tht the vlues given in this poblem sttement e vlid to moe thn two significnt figues.) Is you ppoximte esult gete o smlle thn the exct esult? xplin you nswe. Pictue the Poblem We cn use the definition of λ to find the totl chge of the line of chge nd the expession fo the electic field on the xis of finite line of chge to evlute x t the given loctions long the x xis. In Pt (d) we cn pply Coulomb s lw fo the electic field due to point chge to ppoximte the electic field t x 5 m. () Use the definition of line chge density to expess in tems of λ: λl 8nC (.5nC/m)( 5.m) 7.5nC xpess the electic field on the xis of finite line chge: x ( x ) x k ( x L) (b) Substitute numeicl vlues nd evlute x t x 6. m: x 9 ( ) ( 8.988 N m /C )( 7.5nC) 6.m ( 6.m)( 6.m 5.m) 6N/C (c) Substitute numeicl vlues nd evlute x t x 9. m: x 9 ( ) ( 8.988 N m /C )( 7.5nC) 9.m ( 9.m)( 9.m 5.m) 4.4 N/C (d) Substitute numeicl vlues nd evlute x t x 5 m: x 9 ( ) ( 8.988 N m /C )( 7.5nC) 5m ( 5m)( 5m 5.m).568 mn/c.6mn/c (e) Use Coulomb s lw fo the electic field due to point chge to obtin: k x ( x) x
The lectic Field II: Continuous Chge Distibutions 97 Substitute numeicl vlues nd evlute x (5 m): x 9 ( ) ( 8.988 N m /C )( 7.5nC) 5m ( 5m.5 m).56774mn/c.6 mn/c This esult is bout.% less thn the exct vlue obtined in (d). This suggests tht the line of chge is too long fo its field t distnce of 5 m to be modeled exctly s tht due to point chge. 4 Two infinite non-conducting sheets of chge e pllel to ech othe, with sheet A in the x. m plne nd sheet B in the x +. m plne. Find the electic field in the egion x <. m, in the egion x > +. m, nd between the sheets fo the following situtions. () When ech sheet hs unifom sufce chge density equl to +. μc/m nd (b) when sheet A hs unifom sufce chge density equl to +. μc/m nd sheet B hs unifom sufce chge density equl to. μc/m. (c) Sketch the electic field-line ptten fo ech cse. Pictue the Poblem Let the chge densities on the two pltes be nd nd denote the thee egions of inteest s,, nd. Choose coodinte system in which the positive x diection is to the ight. We cn pply the eqution fo ne n infinite plne of chge nd the supeposition of fields to find the field in ech of the thee egions. () Use the eqution fo ne n infinite plne of chge to expess the field in egion when +. μc/m : + πk iˆ πk iˆ 4πk iˆ Substitute numeicl vlues nd evlute : 9 ( )( ) iˆ 5 8.988 N m /C. C/m (.4 N/C)iˆ 4π μ
98 Chpte Poceed s bove fo egion : ˆ ˆ ˆ ˆ + i i i i π π π π k k k k Poceed s bove fo egion : ( ) ( )i i i i i ˆ N/C.4 ˆ C/m. C m N 8.988 4 ˆ 4 ˆ ˆ 5 9 + + μ π π π π k k k (b) Use the eqution fo ne n infinite plne of chge to expess nd evlute the field in egion when +. μc/m nd. μc/m : ˆ ˆ ˆ ˆ + i i i i π π π π k k k k Poceed s bove fo egion : ( ) ( )i i i i i ˆ N/C.4 ˆ C/m. C m N 8.988 4 ˆ 4 ˆ ˆ 5 9 + + μ π π π π k k k Poceed s bove fo egion : ˆ ˆ ˆ ˆ + i i i i π π π π k k k k (c) The electic field lines fo () nd (b) e shown below: () (b)
The lectic Field II: Continuous Chge Distibutions 99 5 A chge of.75 μc is unifomly distibuted on ing of dius 8.5 cm. Find the electic field stength on the xis t distnces of (). cm, (b).6 cm, nd (c) 4. m fom the cente of the ing. (d) Find the field stength t 4. m using the ppoximtion tht the ing is point chge t the oigin, nd compe you esults fo Pts (c) nd (d). Is you ppoximte esult good one? xplin you nswe. Pictue the Poblem The mgnitude of the electic field on the xis of ing of chge is given by ( z) kx ( z ) x + whee is the chge on the ing nd is the dius of the ing. We cn use this eltionship to find the electic field on the xis of the ing t the given distnces fom the ing. xpess on the xis of ing chge: kx x ( z) + ( z ) () Substitute numeicl vlues nd evlute x fo z. cm: x 9 ( ) ( 8.988 N m /C )(.75μC)(.cm).cm (.cm) + ( 8.5cm) [ ] 4.7 5 N/C (b) Poceed s in () with z.6 cm: x 9 ( ) ( 8.988 N m /C )(.75μC)(.6cm).6cm (.6cm) + ( 8.5cm) [ ]. 6 N/C (c) Poceed s in () with z 4. m: x 9 ( ) ( 8.988 N m /C )(.75μC)( 4.m) 4.m ( 4.m) + ( 8.5cm) [ ].5 N/C (d) Using Coulomb s lw fo the electic field due to point chge, expess z : k z ( z) z Substitute numeicl vlues nd evlute x t z 4. m: z 9 ( ) ( 8.988 N m /C )(.75μC) 4.m ( 4.m).5 N/C
Chpte While this esult gees exctly, to two significnt figues, with the esult obtined in Pt (c), it should be slightly lge becuse the point chge is nee x 4. m thn is the ing of chge. 6 A non-conducting disk of dius R lies in the z plne with its cente t the oigin. The disk hs unifom sufce chge density. Find the vlue of z fo which z /( 4 ). Note tht t this distnce, the mgnitude of the electicfield stength is hlf the electic-field stength t points on the x xis tht e vey close to the disk. Pictue the Poblem The electic field on the xis of disk chge is given by x z π kq. We cn equte this expession nd z ( ) nd z + R solve fo z. xpess the electic field on the xis of disk chge: z π kq z x + R We e given tht: ( ) z 4 qute these expessions: x πk 4 z + R Substituting fo k yields: x π 4 4π z + R Solve fo z to obtin: z R 7 [SSM] A ing tht hs dius lies in the z plne with its cente t the oigin. The ing is unifomly chged nd hs totl chge. Find z on the z xis t () z., (b) z.5, (c) z.7, (d) z, nd (e) z. (f) Use you esults to plot z vesus z fo both positive nd negtive vlues of z. (Assume tht these distnces e exct.)
The lectic Field II: Continuous Chge Distibutions z Pictue the Poblem We cn use z πkq to find the electic z + field t the given distnces fom the cente of the chged ing. () vlute z (.): (.) (b) vlute z (.5): (.5) (c) vlute z (.7): (.7) z z z k(.) (.) + [ ] k.89 k(.5) (.5) + [ ] k.58 k(.7) (.7) + [ ] k.85 (d) vlute z (): (e) vlute z (): k z ( ) ( ) k.54 [ + ] k z k.79 [( ) + ]
Chpte (f) The field long the x xis is plotted below. The z coodintes e in units of z/ nd is in units of k/..4. x. -. -.4 - - - z/ 8 A non-conducting disk of dius lies in the z plne with its cente t the oigin. The disk is unifomly chged nd hs totl chge. Find z on the z xis t () z., (b) z.5, (c) z.7, (d) z, nd (e) z. (f) Use you esults to plot z vesus z fo both positive nd negtive vlues of z. (Assume tht these distnces e exct.) z Pictue the Poblem We cn use z π kq, whee is the dius z + of the disk, to find the electic field on the xis of chged disk. The electic field on the xis of chged disk of dius is given by: z πk z z z + z + () vlute z (.): z (.).4. (.) +
The lectic Field II: Continuous Chge Distibutions (b) vlute z (.5): z (.5).76.5 (.5) + (c) vlute z (.7): z (.7)..7 (.7) + (d) vlute z (): z ( ).46 + (e) vlute z (): z ( ).58 ( ) + The field long the x xis is plotted below. The x coodintes e in units of z/ nd is in units of...6 x..8.4. - - - z/
4 Chpte 9 () Using spedsheet pogm o gphing clculto, mke gph of the electic field stength on the xis of disk tht hs dius. cm nd sufce chge density.5 nc/m. (b) Compe you esults to the esults bsed on the ppoximtion πk (the fomul fo the electic-field stength of unifomly chged infinite sheet). At wht distnce does the solution bsed on ppoximtion diffe fom the exct solution by. pecent? Pictue the Poblem The electic field on the x xis of disk of dius cying z sufce chge density is given by z π k. The electic z + field due to n infinite sheet of chge density is independent of the distnce fom the plne nd is given by sheet πk. () A spedsheet pogm to gph x s function of x is shown below. The fomuls used to clculte the quntities in the columns e s follows: Cell Content/Fomul Algebic Fom B 9.+9 k B4 5. B5. A8 x A9. x +. B8 *PI()*$B$*$B$4*(A8/ z (A8^+$B$5^)^)^.5) π k z + C8 *PI()*$B$*$B$4 πk A B C k 9.+9 N m /C 4 5.- C/m 5. m 6 7 z (z) sheet 8. 8.7 8. 9. 7. 8.. 6.9 8.. 5.46 8..4 4.54 8..5.6 8. 4.6.7 8. 5.7.85 8. 7.65.6 8.
The lectic Field II: Continuous Chge Distibutions 5 74.66.5 8. 75.67.47 8. 76.68.4 8. 77.69.4 8. 78.7.9 8. (b) The following gph shows s function of z. The electic field fom n infinite sheet with the sme chge density is shown fo compison. The mgnitudes diffe by moe thn. pecent fo x. m. 5, N/C 5 _sheet 5.....4.5.6.7 z, m () Show tht the electic-field stength on the xis of ing chge of dius hs mximum vlues t z ±/. (b) Sketch the field stength vesus z fo both positive nd negtive vlues of z. (c) Detemine the mximum vlue of. Pictue the Poblem The electic field on the xis of ing chge s function of distnce z long the xis fom the cente of the ing is given by kz z. We cn show tht it hs its mximum nd minimum vlues t z + ( ) z + nd z by setting its fist deivtive equl to zeo nd solving the esulting eqution fo z. The gph of z will confim tht the mximum nd minimum occu t these coodintes. () The vition of z with z on the kz z z + xis of ing chge is given by: ( )
6 Chpte Diffeentite this expession with espect to z to obtin: d x dz k k d ( z ) z ( z ) d x + + dz k dz ( z + ) ( z + ) ( z + ) z()( z + ) ( z) ( z + ) z ( z + ) k ( z + ) ( z + ) Set this expession equl to zeo fo extem nd simplify: Solving fo z yields: ( z + ) z ( z + ) ( z + ) ( + ) z ( z + ) z, nd z + z z ± s ou cndidtes fo mxim o minim., (b) A plot of the mgnitude of z, in units of k/, vesus z/ follows. This gph shows tht the extem t z ± e, in fct, mxim..4. (z / )... - - - z/
The lectic Field II: Continuous Chge Distibutions 7 (c) vlute z ± nd simplify to obtin the mximum vlue of the mgnitude of z : ± k ± 9 z,mx z ( ) + ± + Remks: Note tht ou esult in Pt (c) confims the mxim obtined gphiclly in Pt (b). A line chge tht hs unifom line chge density λ lies long the x xis fom x x to x x whee x < x. Show tht the x component of the electic field t point on the y-xis is given by x kλ y cosθ cosθ θ tn (x /y), θ tn (x /y) nd y. k k ( ) whee Pictue the Poblem The line chge nd point (, y) e shown in the digm. Also shown is line element of length dx nd the field d its chge poduces t (, y). We cn find d x fom d nd then integte fom x x to x x. xpess the x component of d : d x kλ sinθdx x + y kλ x + y kλx ( x + y ) x x + y dx dx
8 Chpte Integte fom x x to x nd simplify to obtin: ( ) + + + + + + + + y x y y x y y k y x y x k y x k dx y x x k x x x x x λ λ λ λ Fom the digm we see tht: cos y x y + θ o y x tn θ nd cos y x y + θ o y x tn θ Substitute to obtin: [ ] [ ] cos cos cos cos θ θ λ θ θ λ + y k y k x A ing of dius hs chge distibution on it tht vies s λ(θ) λ sin θ, s shown in Figue -9. () Wht is the diection of the electic field t the cente of the ing? (b) Wht is the mgnitude of the field t the cente of the ing? Pictue the Poblem The following digm shows segment of the ing of length ds tht hs chge dq λds. We cn expess the electic field t the cente of the ing due to the chge dq nd then integte this expession fom d θ to π to find the mgnitude of the field in the cente of the ing.
The lectic Field II: Continuous Chge Distibutions 9 () nd (b) The field d t the cente of the ing due to the chge dq is: The mgnitude d of the field t the cente of the ing is: d d x + d y d cosθ iˆ d sinθ ˆj kdq d () Becuse dq λds: The line chge density vies with θ ccoding to λ(θ) λ sin θ : Substitute dθ fo ds: d d d kλds kλ sin θ ds kλ sinθ dθ kλ sinθ dθ Substitute fo d in eqution () to obtin: Integte d fom θ to π nd simplify to obtin: kλ sinθ cosθ dθ d iˆ kλ sin θ dθ ˆj π π kλ k k d iˆ λ sin sin d ˆ π λ θ θ θ θ j ˆj π k λ ˆj (b) The field t the oigin is in the negtive y diection nd its mgnitude is π kλ.
Chpte A line of chge tht hs unifom line chge density λ lies on the x xis fom x to x. Show tht the y component of the electic field t point on the y xis is given by y kλ y y +, y. Pictue the Poblem The line of chge nd the point whose coodintes e (, y) e shown in the digm. Also shown is segment of the line of length dx nd chge dq. The field due to this chge t (, y) is d. We cn find d y fom d nd then integte fom x to x to find the y component of the electic field t point on the y xis. d y θ (, y) dq dx x () xpess the mgnitude of the field d due to chge dq of the element of length dx: kdq d whee x + y Becuse dq λdx : d kλdx x + y xpess the y component of d: d y kλ x y + cosθ dx Refe to the digm to expess y cosθ cosθ in tems of x nd y: x + y Substitute fo cosθ in the kλy d y expession fo d y to obtin: ( x + y ) Integte fom x to x nd simplify to obtin: dx y kλ y dx kλy ( x + y ) y x + y y + y x kλ
The lectic Field II: Continuous Chge Distibutions 4 Clculte the electic field distnce z fom unifomly chged infinite flt non-conducting sheet by modeling the sheet s continuum of infinite stight lines of chge. Pictue the Poblem The field due to line of chge is given by λ () whee is the π pependicul distnce to the line. The digm shows point P, t which we will clculte the electic field due continuum of infinite stight nonconducting lines of chge, nd few of the lines of chge. P is distnce L fom the plne nd the oigin of the coodinte system is diectly below P. Note tht the hoizontl components of the field t P, by symmety, dd up to zeo. Hence we need only find the sum of ll the z components of the field. Becuse the hoizontl components of the electic field dd up to zeo, the esultnt field is given by: xpess the field due to n infinite line of chge: The sufce chge density of the plne nd the line chge density of the chged ings λ e elted: Substitute fo d λ to obtin: dy () d θ y y π π () d cosθ () dλ d(), whee is the π pependicul distnce to the line of chge. dλ dy d () dy π P x Substituting fo d() in eqution () yields: Refeing to the digm, note tht: π π π dy cosθ y x tnθ dy xsec θ dθ
Chpte Substitute fo dy in the expession fo to obtin: π π π xsec θ cosθ dθ Becuse x cosθ : π π π sec θ cos θ dθ π π π dθ Integting this expession yields: π θ π π d π π 5 [SSM] Clculte the electic field distnce z fom unifomly chged infinite flt non-conducting sheet by modeling the sheet s continuum of infinite cicul ings of chge. Pictue the Poblem The field t point on the xis of unifomly chged ing lies long the xis nd is given by qution -8. The digm shows one ing of the continuum of cicul ings of chge. The dius of the ing is nd the distnce fom its cente to the field point P is x. The ing hs unifomly distibuted chge. The esultnt electic field t P is the sum of the fields due to the continuum of cicul ings. Note tht, by symmety, the hoizontl components of the electic field cncel. d x P d xpess the field of single unifomly chged ing with chge nd dius on the xis of the ing t distnce x wy fom the plne of the ing: xi ˆ kx, whee x x + ( ) Substitute dq fo nd d x fo x to kxdq d x x + obtin: ( )
The lectic Field II: Continuous Chge Distibutions The esultnt electic field t P is the sum of the fields due to ll the cicul ings. Integte both sides to clculte the esultnt field fo the entie plne. The field point emins fixed, so x is constnt: To evlute this integl we chnge integtion vibles fom q to. The chge dq da whee da π d is the e of ing of dius nd width d: To integte this expession, letu x +. Then: Noting tht when, u x, substitute nd simplify to obtin: kxdq dq kx ( ) x + ( x + ) dq π d so π kx x + π kx ( ) d d ( x + ) du x + u o d udu ( d) d u π kx du π kx u x x u du vluting the integl yields: π kx πk u x 6 A thin hemispheicl shell of dius R hs unifom sufce chge. Find the electic field t the cente of the bse of the hemispheicl shell. Pictue the Poblem Conside the ing with its xis long the z diection shown in the digm. Its dius is z cosθ nd its width is dθ. We cn use the eqution fo the field on the xis of ing chge nd then integte to expess the field t the cente of the hemispheicl shell. d cosθ z θ sinθ dθ dθ y x
4 Chpte xpess the field on the xis of d the ing chge: ( sin θ + cos θ ) kzdq whee z cosθ kzdq xpess the chge dq on the ing: dq da ( π sinθ ) π sinθdθ Substitute to obtin: k( cosθ ) dθ π sinθdθ d πk sinθ cosθdθ Integting d fom θ to π/ yields: πk sinθ cosθdθ πk π π [ sin θ ] πk Guss s Lw 7 A sque tht hs -cm-long edges is centeed on the x xis in egion whee thee exists unifom electic field given by (. kn/c)iˆ. () Wht is the electic flux of this electic field though the sufce of sque if the noml to the sufce is in the +x diection? (b) Wht is the electic flux though the sme sque sufce if the noml to the sufce mkes 6º ngle with the y xis nd n ngle of 9 with the z xis? Pictue the Poblem The definition of electic flux isφ nˆda. We cn S pply this definition to find the electic flux though the sque in its two oienttions. () Apply the definition of φ to find the flux of the field when the sque is pllel to the yz plne: φ (.kn/c) S (.kn/c) (.kn/c)(.m). N m iˆ ida ˆ S da /C
(b) Poceed s in () with ˆ n ˆ cos : The lectic Field II: Continuous Chge Distibutions 5 φ (. kn/c) S i (. kn/c) (. kn/c)(.m) 7 N m /C cos da cos S da cos 8 A single point chge (q +. μc) is fixed t the oigin. An imginy spheicl sufce of dius. m is centeed on the x xis t x 5. m. () Sketch electic-field lines fo this chge (in two dimensions) ssuming twelve eqully-spced field lines in the xy plne leve the chge loction, with one of the lines in the +x diection. Do ny lines ente the spheicl sufce? If so, how mny? (b) Do ny lines leve the spheicl sufce? If so, how mny? (c) Counting the lines tht ente s negtive nd the ones tht leve s positive, wht is the net numbe of field lines tht penette the spheicl sufce? (d) Wht is the net electic flux though this spheicl sufce? Detemine the Concept We must show the twelve electic field lines oiginting t q nd, in the bsence of othe chges, dilly symmetic with espect to the loction of q. While we e dwing twelve lines in this poblem, the numbe of lines tht we dw is lwys, by geement, in popotion to the mgnitude of q. () The sketch of the field lines nd of the spheicl sufce is shown in the digm to the ight. q Given the numbe of field lines dwn fom q, lines ente the spheicl sufce. Hd we chosen to dw 4 field lines, 6 would hve enteed the spheicl sufce. (b) Thee lines leve the spheicl sufce. (c) Becuse the thee lines tht ente the spheicl sufce lso leve the spheicl sufce, the net numbe of field lines tht pss though the sufce is zeo. (d) Becuse s mny field lines leve the spheicl sufce s ente it, the net flux is zeo.
6 Chpte 9 [SSM] An electic field is given by sign( x) ( N/C)iˆ, whee sign(x) equls if x <, if x, nd + if x >. A cylinde of length cm nd dius 4. cm hs its cente t the oigin nd its xis long the x xis such tht one end is t x + cm nd the othe is t x cm. () Wht is the electic flux though ech end? (b) Wht is the electic flux though the cuved sufce of the cylinde? (c) Wht is the electic flux though the entie closed sufce? (d) Wht is the net chge the cylinde? Pictue the Poblem The field t both cicul fces of the cylinde is pllel to the outwd vecto noml to the sufce, so the flux is just A. Thee is no flux though the cuved sufce becuse the noml to tht sufce is pependicul to. The net flux though the closed sufce is elted to the net chge by Guss s lw. () Use Guss s lw to clculte the flux though the ight cicul sufce: Apply Guss s lw to the left cicul sufce: φ φ ight left ight nˆ ight ( N/C) iˆ iˆ ( π )(.4 m).5 N m left nˆ left A A /C ( N/C) iˆ ( iˆ )( π )(.4 m).5 N m /C (b) Becuse the field lines e pllel to the cuved sufce of the cylinde: φ cuved
The lectic Field II: Continuous Chge Distibutions 7 (c) xpess nd evlute the net flux though the entie cylindicl sufce: φ net φ + φ + φ ight left.5 N m /C +.5 N m /C +. N m /C cuved (d) Apply Guss s lw to obtin: φ net 4πk φnet 4πk Substitute numeicl vlues nd evlute : π 9 4 ( 8.988 N m /C ).7. N m C /C Ceful mesuement of the electic field t the sufce of blck box indictes tht the net outwd electic flux though the sufce of the box is 6. kn m /C. () Wht is the net chge the box? (b) If the net outwd electic flux though the sufce of the box wee zeo, could you conclude tht thee wee no chges the box? xplin you nswe. Pictue the Poblem We cn use Guss s lw in tems of to find the net chge the box. () Apply Guss s lw in tems of φnet φnet to find the net chge the box: Substitute numeicl vlues nd evlute : 8.854 C N m kn m 6. C 5. 8 C (b) You cn only conclude tht the net chge is zeo. Thee my be n equl numbe of positive nd negtive chges pesent the box. A point chge (q +. μc) is t the cente of n imginy sphee tht hs dius equl to.5 m. () Find the sufce e of the sphee. (b) Find the mgnitude of the electic field t ll points on the sufce of the sphee. (c) Wht is the flux of the electic field though the sufce of the sphee? (d) Would you nswe to Pt (c) chnge if the point chge wee moved so tht it ws the sphee but not t its cente? (e) Wht is the flux of the electic field though the sufce of n imginy cube tht hs.-m-long edges nd encloses the sphee?
8 Chpte Pictue the Poblem We cn pply Guss s lw to find the flux of the electic field though the sufce of the sphee. () Use the fomul fo the sufce e of sphee to obtin: A 4π 4π.4m (.5 m).4 m (b) Apply Coulomb s lw to find : 4π q 4π 7.9 4 N/C.μC ( 8.854 C /N m )(.5m) 7.9 4 N/C (c) Apply Guss s lw to obtin: nda ˆ da S S 4 ( 7.9 N/C)(.4 m ) φ.6 5 N m /C (d) No. The flux though the sufce is independent of whee the chge is locted the sphee. (e) Becuse the cube encloses the sphee, the flux though the sufce of the sphee will lso be the flux though the cube: φ cube.6 5 N m /C Wht is the electic flux though one side of cube tht hs single point chge of. μc plced t its cente? HINT: You do not need to integte ny equtions to get the nswe. Pictue the Poblem The flux though the cube is given by φnet, whee is the chge t the cente of the cube. The flux though one side of the cube is one-sixth of the totl flux though the cube. The flux though one side of the cube is one-sixth of the totl flux though the cube: φ fce 6 φ tot 6
The lectic Field II: Continuous Chge Distibutions 9 Substitute numeicl vlues nd evlute φ : fce φ fce. μc C 6 8.854 N m 5.65 4 N m C [SSM] A single point chge is plced t the cente of n imginy cube tht hs -cm-long edges. The electic flux out of one of the cube s sides is.5 kn m /C. How much chge is t the cente? Pictue the Poblem The net flux though the cube is given by φnet whee is the chge t the cente of the cube., The flux though one side of the cube is one-sixth of the totl flux though the cube: φ fces 6 φ net 6 Solving fo yields: 6 φ fces Substitute numeicl vlues nd evlute : C kn m 6 8.854.5 N m C 79.7 nc 4 Becuse the fomuls fo Newton s lw of gvity nd fo Coulomb s lw hve the sme invese-sque dependence on distnce, fomul nlogous to the fomul fo Guss s lw cn be found fo gvity. The gvittionl field g t loction is the foce pe unit mss on test mss m plced t tht loction. Then, fo point mss m t the oigin, the gvittionl field g t some position ( ) is g ( Gm )ˆ. Compute the flux of the gvittionl field though spheicl sufce of dius R centeed t the oigin, nd veify tht the gvittionl nlog of Guss s lw is φ net 4πGm. Pictue the Poblem We ll define the flux of the gvittionl field in mnne tht is nlogous to the definition of the flux of the electic field nd then substitute fo the gvittionl field nd evlute the integl ove the closed spheicl sufce. Define the gvittionl flux s: φ g nˆda g S
Chpte Substitute fo g nd evlute the integl to obtin: φ net Gm Gm Gm S 4πGm S da ˆ nˆ da ( 4π ) 5 An imginy ight cicul cone (Figue -4) tht hs bse ngle θ nd bse dius R is in chge fee egion tht hs unifom electic field (field lines e veticl nd pllel to the cone s xis). Wht is the tio of the numbe of field lines pe unit e penetting the bse to the numbe of field lines pe unit e penetting the conicl sufce of the cone? Use Guss's lw in you nswe. (The field lines in the figue e only epesenttive smple.) Pictue the Poblem Becuse the cone encloses no chge, we know, fom Guss s lw, tht the net flux of the electic field though the cone s sufce is zeo. Thus, the numbe of field lines penetting the cuved sufce of the cone must equl the numbe of field lines penetting the bse nd the enteing flux must equl the exiting flux. The flux penetting the bse of the cone is given by: R φ enteing A bse θ nˆ θ The flux penetting the cuved sufce of the cone is given by: φ exiting nda ˆ S S cosθ da quting the fluxes nd simplifying yields: A bse cosθ da S ( cosθ ) A cuved sufce The tio of the density of field lines is: A A bse cuved sufce cosθ 6 In the tmosphee nd t n ltitude of 5 m, you mesue the electic field to be 5 N/C diected downwd nd you mesue the electic field to be 7 N/C diected downwd t n ltitude of 4 m. Clculte the volume
The lectic Field II: Continuous Chge Distibutions chge density of the tmosphee in the egion between ltitudes of 5 m nd 4 m, ssuming it to be unifom. (You my neglect the cuvtue of th. Why?) Pictue the Poblem We ll model this potion of th s tmosphee s though it is cylinde with coss-sectionl e A nd height h. Becuse the electic flux inceses with ltitude, we cn conclude tht thee is chge the cylindicl egion nd use Guss s lw to find tht chge nd hence the chge density of the tmosphee in this egion. The definition of volume chge density is: xpess the chge cylinde of bse e A nd height h fo chge density ρ: Tking upwd to be the positive diection, pply Guss s lw to the chge in the cylinde: ρ V ρah ( A A) ( A A) h whee we ve tken ou zeo t 5 m bove the sufce of flt th. h Substitute to obtin: ( A A) ( ) Substitute numeicl vlues nd evlute ρ: ρ ρ h h ( 5 N/C 7 N/C)( 8.854 C /N m ) 4m 5m Ah. h C/m whee we ve been ble to neglect the cuvtue of th becuse the mximum height of 4 m is ppoximtely.6% of the dius of th. Guss s Lw Applictions in Spheicl Symmety Situtions 7 A thin non-conducting spheicl shell of dius R hs totl chge q tht is unifomly distibuted on its sufce. A second, lge thin nonconducting spheicl shell of dius R tht is coxil with the fist hs chge q tht is unifomly distibuted on its sufce. () Use Guss s lw to obtin expessions fo the electic field in ech of the thee egions: < R, R < < R, nd > R. (b) Wht should the tio of the chges q /q nd the eltive signs fo q nd q be fo the electic field to be zeo thoughout the egion > R? (c) Sketch the electic field lines fo the sitution in Pt (b) when q is positive.
Chpte Pictue the Poblem To find n in these thee egions we cn choose Gussin sufces of ppopite dii nd pply Guss s lw. On ech of these sufces, is constnt nd Guss s lw eltes to the totl chge the sufce. () Use Guss s lw to find the electic field in the egion < R : S nd n da R < A vecto. ˆ whee ˆ is unit dil Becuse : <R Apply Guss s lw in the egion < < R R < < R : ( 4π ) Using Guss s lw, find the q ˆ R > R electic field in the egion > R : ( 4π ) q + q ˆ k kq ˆ ( q + q ) ˆ (b) Set >R to obtin: q q + q q (c) The electic field lines fo the sitution in (b) with q positive is shown to the ight. 8 A spheicl shell of dius 6. cm cies unifom sufce chge density of A non-conducting thin spheicl shell of dius 6. cm hs unifom sufce chge density of 9. nc/m. () Wht is the totl chge on the shell? Find the electic field t the following distnces fom the sphee s cente: (b). cm, (c) 5.9 cm, (d) 6. cm, nd (e). cm. Pictue the Poblem We cn use the definition of sufce chge density nd the fomul fo the e of sphee to find the totl chge on the shell. Becuse the chge is distibuted unifomly ove spheicl shell, we cn choose spheicl
The lectic Field II: Continuous Chge Distibutions Gussin sufce nd pply Guss s lw to find the electic field s function of the distnce fom the cente of the spheicl shell. () Using the definition of sufce chge density, elte the chge on the sphee to its e: Substitute numeicl vlues nd evlute : A 4π 4π ( 9. nc/m )(.6m).47nC.47 nc Apply Guss s lw to spheicl sufce of dius tht is concentic the spheicl shell to obtin: S n da 4π n Solving fo n yields: (b) The chge sphee whose dius is. cm is zeo nd hence: (c) The chge sphee whose dius is 5.9 cm is zeo nd hence: k n 4π n (.cm) n ( 5.9cm) (d) The chge sphee whose dius is 6. cm is.47 nc nd hence: 9 ( ) ( 8.988 N m /C )(.47nC) 6.cm (.6m) n 98N/C (e) The chge sphee whose dius is. cm is.47 nc nd hence: 9 ( ) ( 8.988 N m /C )(.47 nc) cm (.m) n 66 N/C 9 [SSM] A non-conducting sphee of dius 6. cm hs unifom volume chge density of 45 nc/m. () Wht is the totl chge on the sphee? Find the electic field t the following distnces fom the sphee s cente: (b). cm, (c) 5.9 cm, (d) 6. cm, nd (e). cm.
4 Chpte Pictue the Poblem We cn use the definition of volume chge density nd the fomul fo the volume of sphee to find the totl chge of the sphee. Becuse the chge is distibuted unifomly thoughout the sphee, we cn choose spheicl Gussin sufce nd pply Guss s lw to find the electic field s function of the distnce fom the cente of the sphee. () Using the definition of volume chge density, elte the chge on the sphee to its volume: Substitute numeicl vlues nd evlute : ρ V π 4 4 πρ ( 45 nc/m )(.6 m).47 nc.47 nc Apply Guss s lw to spheicl sufce of dius < R tht is concentic with the spheicl shell to obtin: S n da 4π n Solving fo n yields: Becuse the chge distibution is unifom, we cn find the chge the Gussin sufce by using the definition of volume chge density to estblish the popotion: k n 4π V V' whee V is the volume of the Gussin sufce. Solve fo to obtin: V' V R Substitute fo to obtin: 4π k R ( < R) n (b) vlute n t. cm: 9 ( ) ( 8.988 N m /C )(.47nC).cm (.m) 9 N/C (.6m) n
The lectic Field II: Continuous Chge Distibutions 5 (c) vlute n t 5.9 cm: 9 ( ) ( 8.988 N m /C )(.47nC) 5.9cm (.59m).kN/C (.6m) n Apply Guss s lw to the Gussin k k 4π n n sufce with > R: (d) vlute n t 6. cm: 9 ( ) ( 8.988 N m /C )(.47nC) 6.cm (.6m) n 98N/C (e) vlute n t. cm: 9 ( ) ( 8.988 N m /C )(.47nC).cm (.m) n 66 N/C 4 Conside the solid conducting sphee nd the concentic conducting spheicl shell in Figue -4. The spheicl shell hs chge 7. The solid sphee hs chge +. () How much chge is on the oute sufce nd how much chge is on the inne sufce of the spheicl shell? (b) Suppose metl wie is now connected between the solid sphee nd the shell. Afte electosttic equilibium is e-estblished, how much chge is on the solid sphee nd on ech sufce of the spheicl shell? Does the electic field t the sufce of the solid sphee chnge when the wie is connected? If so, in wht wy? (c) Suppose we etun to the conditions in Pt (), with + on the solid sphee nd 7 on the spheicl shell. We next connect the solid sphee to gound with metl wie, nd then disconnect it. Then how much totl chge is on the solid sphee nd on ech sufce of the spheicl shell? Detemine the Concept The chges on conducting sphee, in esponse to the epulsive Coulomb foces ech expeiences, will septe until electosttic equilibium conditions exit. The use of wie to connect the two sphees o to gound the oute sphee will cuse dditionl edistibution of chge. () Becuse the oute sphee is conducting, the field in the thin shell must vnish. Theefoe,, unifomly distibuted, esides on the inne sufce, nd 5, unifomly distibuted, esides on the oute sufce.
6 Chpte (b) Now thee is no chge on the inne sufce nd 5 on the oute sufce of the spheicl shell. The electic field just outside the sufce of the inne sphee chnges fom finite vlue to zeo. (c) In this cse, the 5 is dined off, leving no chge on the oute sufce nd on the inne sufce. The totl chge on the oute sphee is then. 4 A non-conducting solid sphee of dius. cm hs unifom volume chge density. The mgnitude of the electic field t. cm fom the sphee s cente is.88 N/C. () Wht is the sphee s volume chge density? (b) Find the mgnitude of the electic field t distnce of 5. cm fom the sphee s cente. Pictue the Poblem () We cn use the definition of volume chge density, in conjunction with qution -8, to find the sphee s volume chge density. (b) We cn use qution -8b, in conjunction with ou esult fom Pt (), to find the electic field t distnce of 5. cm fom the solid sphee s cente. () The solid sphee s volume chge density is the tio of its chge to its volume: () V ρ 4 πr Fo R, qution -8 gives the electic field t distnce fom the cente of the sphee: () 4π Solving fo yields: 4π Substitute fo () nd simplify to obtin: in eqution 4π ρ 4 πr R Substitute numeicl vlues nd evlute ρ: ( C /N m )(.88 N/C)(. cm) 8.854 ρ.μc/m (. cm).997μc/m (b) Fo R, the electic field t distnce fom the cente of the sphee is given by: () 4π R
The lectic Field II: Continuous Chge Distibutions 7 xpess fo R: ρv sphee whose diusis 4 π ρ Substituting fo in eqution () nd simplifying yields: 4π 4 π R 4 ρ ρ R Substitute numeicl vlues nd evlute (5. cm): 4 ( ) (.997μC/m )( 5. cm). cm ( 8.854 C /N m )(. cm) 5 47 N/C 4 A non-conducting solid sphee of dius R hs volume chge density tht is popotionl to the distnce fom the cente. Tht is, ρ Afo R, whee A is constnt. () Find the totl chge on the sphee. (b) Find the expessions fo the electic field the sphee ( < R) nd outside the sphee ( > R). (c) Sketch the mgnitude of the electic field s function of the distnce fom the sphee s cente. Pictue the Poblem We cn find the totl chge on the sphee by expessing the chge dq in spheicl shell nd integting this expession between nd R. By symmety, the electic fields must be dil. To find the chged sphee we choose spheicl Gussin sufce of dius < R. To find outside the chged sphee we choose spheicl Gussin sufce of dius > R. On ech of these sufces, is constnt. Guss s lw then eltes to the totl chge the sufce. () xpess the chge dq in shell of thickness d nd volume 4π d: Integte this expession fom to R to find the totl chge on the sphee: (b) Apply Guss s lw to spheicl sufce of dius > R tht is concentic with the nonconducting sphee to obtin: dq 4π ρd 4π 4πA d ( A) d 4 R 4 [ πa ] π R 4π A d AR S da 4π
8 Chpte Solving fo yields: ( > R) 4π k kaπr 4 4 AR 4 Apply Guss s lw to spheicl sufce of dius < R tht is concentic with the nonconducting sphee to obtin: S da 4π Solve fo to obtin: ( < R) 4π A 4 4 πa 4π (c) The following gph of vesus /R, with in units of A/(4 ), ws plotted using spedsheet pogm...8.6.4....5..5..5. /R Remks: Note tht the esults fo () nd (b) gee t R. 4 [SSM] A sphee of dius R hs volume chge density ρ B/ fo < R, whee B is constnt nd ρ fo > R. () Find the totl chge on the sphee. (b) Find the expessions fo the electic field nd outside the chge distibution (c) Sketch the mgnitude of the electic field s function of the distnce fom the sphee s cente. Pictue the Poblem We cn find the totl chge on the sphee by expessing the chge dq in spheicl shell nd integting this expession between nd
The lectic Field II: Continuous Chge Distibutions 9 R. By symmety, the electic fields must be dil. To find the chged sphee we choose spheicl Gussin sufce of dius < R. To find outside the chged sphee we choose spheicl Gussin sufce of dius > R. On ech of these sufces, is constnt. Guss s lw then eltes to the totl chge the sufce. () xpess the chge dq in shell of thickness d nd volume 4π d: Integte this expession fom to R to find the totl chge on the sphee: dq 4π ρd 4π 4πBd R 4πB d πbr B [ πb ] d R (b) Apply Guss s lw to spheicl sufce of dius > R tht is concentic with the nonconducting sphee to obtin: S da o 4π Solving fo yields: ( > R) 4π k kπbr BR Apply Guss s lw to spheicl sufce of dius < R tht is concentic with the nonconducting sphee to obtin: S da 4π Solving fo yields: ( < R) 4π B πb 4π
Chpte (c) The following gph of vesus /R, with in units of B/( ), ws plotted using spedsheet pogm....8.6.4....5..5..5. /R Remks: Note tht ou esults fo () nd (b) gee t R. 44 A sphee of dius R hs volume chge density ρ C/ fo < R, whee C is constnt nd ρ fo > R. () Find the totl chge on the sphee. (b) Find the expessions fo the electic field nd outside the chge distibution (c) Sketch the mgnitude of the electic field s function of the distnce fom the sphee s cente. Pictue the Poblem We cn find the totl chge on the sphee by expessing the chge dq in spheicl shell nd integting this expession between nd R. By symmety, the electic fields must be dil. To find the chged sphee we choose spheicl Gussin sufce of dius < R. To find outside the chged sphee we choose spheicl Gussin sufce of dius > R. On ech of these sufces, is constnt. Guss s lw then eltes to the totl chge the sufce. () xpess the chge dq in shell of thickness d nd volume 4π d: Integte this expession fom to R to find the totl chge on the sphee: dq C π ρd 4π d 4πCd 4 R R [ 4πC] πcr 4πC d 4
The lectic Field II: Continuous Chge Distibutions (b) Apply Guss s lw to spheicl sufce of dius > R tht is concentic with the nonconducting sphee to obtin: S da 4π Solving fo yields: ( > R) k 4π k4πcr CR Apply Guss s lw to spheicl sufce of dius < R tht is concentic with the nonconducting sphee to obtin: S da o 4π Solving fo yields: ( < R) 4π C 4πC 4π (c) The following gph of vesus /R, with in units of ( R) plotted using spedsheet pogm. C /, ws 8 6 4..5..5..5. /R 45 A non-conducting spheicl shell of inne dius R nd oute dius R hs unifom volume chge density ρ. () Find the totl chge on the shell. (b) Find expessions fo the electic field eveywhee.
Chpte Pictue the Poblem By symmety, the electic fields esulting fom this chge distibution must be dil. To find fo < R we choose spheicl Gussin sufce of dius < R. To find fo R < < R we choose spheicl Gussin sufce of dius R < < R. To find fo > R we choose spheicl Gussin sufce of dius > R. On ech of these sufces, is constnt. Guss s lw then eltes to the totl chge the sufce. () The chge in n infinitesiml spheicl shell of dius nd thickness d is: d ρ dv 4πρ d Integte d fom R to to find the totl chge in the spheicl shell in the intevl R < < R : 4πρ R 4πρ 4πC d ( R ) R (b) Apply Guss s lw to spheicl sufce of dius tht is concentic with the nonconducting spheicl shell to obtin: S da 4π Solving fo yields: k () 4π vlute ( < R ): ( < R ) 4π k becuse ρ( < R ) nd, theefoe,. vlute (R < < R ): ( R < < R ) k 4πkρ ρ ( R R ) ( R R )
The lectic Field II: Continuous Chge Distibutions Fo > R : nd Remks: Note tht is continuous t R. 4 πρ ( R R ) 4πkρ ( > R ) ( R R ) ρ ( R R ) Guss s Lw Applictions in Cylindicl Symmety Situtions 46 Fo you senio poject you e in chge of designing Geige tube fo detecting dition in the nucle physics lbotoy. This instument will consist of long metl cylindicl tube tht hs long stight metl wie unning down its centl xis. The dimete of the wie is to be.5 mm nd the dimete of the tube will be 4. cm. The tube is to be filled with dilute gs in which electicl dischge (bekdown) occus when the electic field eches 5.5 6 N/C. Detemine the mximum line chge density on the wie if bekdown of the gs is not to hppen. Assume tht the tube nd the wie e infinitely long. Pictue the Poblem The electic field of line chge of infinite length is given λ by, whee is the distnce fom the cente of the line of chge nd π λ is the line chge density of the wie. The electic field of line chge of infinite length is given by: λ π Becuse vies invesely with, its mximum vlue occus t the sufce of the wie whee R, the dius of the wie: mx λ R π Solving fo λ yields: λ π R mx Substitute numeicl vlues nd evlute λ: λ π 8.854 C N m N C 6 (.5 mm) 5.5 76.5 nc/m
4 Chpte 47 In Poblem 54, suppose ionizing dition poduces n ion nd n electon t distnce of. cm fom the long xis of the centl wie of the Geige tube. Suppose tht the centl wie is positively chged nd hs line chge density equl to 76.5 pc/m. () In this cse, wht will be the electon s speed s it impcts the wie? (b) ulittively, how will the electon s speed compe to tht of the ion s finl speed when it impcts the outside cylinde? xplin you esoning. Pictue the Poblem Becuse the inwd foce on the electon inceses s its distnce fom the wie deceses, we ll need to integte the net electic foce cting on the electon to obtin n expession fo its speed s function of its distnce fom the wie in the Geige tube. () The foce the electon expeiences is the dil component of the foce on the electon nd is the poduct of its chge nd the dil component of the electic field due to the positively chged centl wie: The dil electic field due to the chged wie is given by: F e e, λ π Substituting fo yields: eλ F e, whee the minus π sign indictes tht the foce cting on the electon is dilly inwd. Apply Newton s nd lw to the electon to obtin: Septing vibles yields: xpess the integl of this eqution to obtin: Integting yields: eλ π dv m dt dv d m d dt eλ d vdv m π vf dv d m dt d dv mv d eλ d vdv whee the πm lowe limit on the left-hnd side is zeo becuse the electon is initilly t est. eλ vf ln πm
The lectic Field II: Continuous Chge Distibutions 5 Solve fo v f to obtin: v f eλ ln πm Substitute numeicl vlues nd evlute v f : v f π 9 (.6 C) ( 9.9 kg).46 6 m/s pc 76.5 m C 8.854 N m. m.5 m ln.5 mm (b) The positive ion is cceleted dilly outwd nd will impct the tube insted of the wie. Becuse of its much lge mss, the impct speed of the ion will be much less thn the impct speed of the electon. 48 Show tht the electic field due to n infinitely long, unifomly chged thin cylindicl shell of dius hving sufce chge density is given by the following expessions: fo R < nd R R ( ) fo R >. Pictue the Poblem Fom symmety, the field in the tngentil diection must vnish. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the infinitely long, unifomly chged cylindicl shell. Apply Guss s lw to the cylindicl sufce of dius nd length L tht is concentic with the infinitely long, unifomly chged cylindicl shell: n da S o πl R whee we ve neglected the end es becuse no thee is no flux though them. Solve fo R : R πl k L Fo < R, nd: ( < R) R
6 Chpte Fo > R, λl nd: R ( > R) kλl kλ k L R ( πr ) 49 A thin cylindicl shell of length m nd dius 6. cm hs unifom sufce chge density of 9. nc/m. () Wht is the totl chge on the shell? Find the electic field t the following dil distnces fom the long xis of the cylinde. (b). cm, (c) 5.9 cm, (d) 6. cm, nd (e). cm. (Use the esults of Poblem 48.) Pictue the Poblem We cn use the definition of sufce chge density to find the totl chge on the shell. Fom symmety, the electic field in the tngentil diection must vnish. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the unifomly chged cylindicl shell. () Using its definition, elte the sufce chge density to the totl chge on the shell: Substitute numeicl vlues nd evlute : A πrl π (.6m)( m)( 9. nc/m ) 679nC (b) Fom Poblem 48 we hve, fo. cm: (c) Fom Poblem 48 we hve, fo 5.9 cm: (.cm) ( 5.9cm) (d) Fom Poblem 48 we hve, fo 6. cm: nd () R ( ) ( 6 9.nC/m )(.6m).cm ( 8.854 C /N m )(.6m).kN/C
The lectic Field II: Continuous Chge Distibutions 7 (e) Fom Poblem 48 we hve, fo. cm: ( ) ( 9.nC/m )(.6m).cm ( 8.854 C /N m )(.m) 6 N/C 5 An infinitely long non-conducting solid cylinde of dius hs unifom volume chge density of ρ. Show tht the electic field is given by the following expessions: R ρ R ( ) fo R < nd R ρ ( R) fo R >, whee R is the distnce fom the long xis of the cylinde. Pictue the Poblem Fom symmety, the field tngent to the sufce of the cylinde must vnish. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the infinitely long nonconducting cylinde. Apply Guss s lw to cylindicl sufce of dius nd length L tht is concentic with the infinitely long nonconducting cylinde: n da S o πl R whee we ve neglected the end es becuse thee is no flux though them. Solving fo R yields: R πl k L ρ ( ) V ρ ( π L) xpess fo < R: Substitute to obtin: ( πρ L ) k ρ R ( < R) L o, becuse λ ρπr, λ π R ( < R) R ρ ( ) V ρ ( πr L) xpess fo > R:
8 Chpte Substitute fo to obtin: ( πρ LR ) k ρr R ( > R) L o, becuse λ ρπr ( > R) λ π R 5 [SSM] A solid cylinde of length m nd dius 6. cm hs unifom volume chge density of nc/m. () Wht is the totl chge of the cylinde? Use the fomuls given in Poblem 5 to clculte the electic field t point equidistnt fom the ends t the following dil distnces fom the cylindicl xis: (b). cm, (c) 5.9 cm, (d) 6. cm, nd (e). cm. Pictue the Poblem We cn use the definition of volume chge density to find the totl chge on the cylinde. Fom symmety, the electic field tngent to the sufce of the cylinde must vnish. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the unifomly chged cylinde. () Use the definition of volume chge density to expess the totl chge of the cylinde: tot ρ V ρ ( πr L) Substitute numeicl vlues to obtin: tot π ( nc/m )(.6 m)( m) 679nC (b) Fom Poblem 5, fo < R, we hve: ρ () Fo. cm: ( ) ( nc/m )(.m).cm ( C /N m ) 8.854 9 N/C (c) Fo 5.9 cm: ( ) ( nc/m )(.59m).9cm 5 ( C /N m ) 8.854.kN/C
The lectic Field II: Continuous Chge Distibutions 9 Fom Poblem 5, fo > R, we hve: () ρr (d) Fo 6. cm: ( ) ( 6 nc/m )(.6m).cm ( 8.854 C /N m )(.6m). kn/c (e) Fo. cm: ( ) ( nc/m )(.6m).cm ( 8.854 C /N m )(.m) 6 N/C 5 Conside two infinitely long, coxil thin cylindicl shells. The inne shell hs dius nd hs unifom sufce chge density of, nd the oute shell hs dius nd hs unifom sufce chge density of. () Use Guss s lw to find expessions fo the electic field in the thee egions: R <, < R <, nd R >, whee R is the distnce fom the xis. (b) Wht is the tio of the sufce chge densities / nd thei eltive signs if the electic field is to be zeo eveywhee outside the lgest cylinde? (c) Fo the cse in Pt (b), wht would be the electic field between the shells? (d) Sketch the electic field lines fo the sitution in Pt (b) if is positive. Pictue the Poblem Fom symmety; the field tngent to the sufces of the shells must vnish. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the infinitely long, unifomly chged cylindicl shells. () Apply Guss s lw to the cylindicl sufce of dius nd length L tht is concentic with the infinitely long, unifomly chged cylindicl shell: n da πl R S whee we ve neglected the end es becuse thee is no flux though them. Solving fo R yields: R k () L Fo < R, nd: ( < R ) R xpess fo R < < R : A π R L
4 Chpte Substitute in eqution () to obtin: R ( R < < R ) ( π R L) k L R xpess fo > R: A + A π R L + π R L Substitute in eqution () to obtin: R ( > R ) ( π R L + π R L) k L R + R (b) Set fo > R to obtin: R + R R R (c) Becuse the electic field is detemined by the chge the Gussin sufce, the field unde these conditions would be s given bove: ( R < < R ) R R (d) Becuse is positive, the field lines e diected s shown to the ight: 5 Figue -4 shows potion of n infinitely long, concentic cble in coss section. The inne conducto hs chge of 6. nc/m nd the oute conducto hs no net chge. () Find the electic field fo ll vlues of R, whee R is the pependicul distnce fom the common xis of the cylindicl system. (b) Wht e the sufce chge densities on the nd the outside sufces of the oute conducto? Pictue the Poblem The electic field is diected dilly outwd. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the infinitely long, unifomly chged cylindicl shell.
The lectic Field II: Continuous Chge Distibutions 4 () Apply Guss s lw to cylindicl sufce of dius nd length L tht is concentic with the inne conducto: S n da πl R whee we ve neglected the end es becuse thee is no flux though them. Solving fo R yields: R k () L Fo <.5 cm, nd: ( <.5cm ) R Letting R.5 cm, expess fo.5 cm < < 4.5 cm: λl πrl Substitute in eqution () to obtin: k R (.5cm < < 4.5cm) L kλ Substitute numeicl vlues nd evlute n (.5 cm < < 4.5 cm): R ( λl) ( ) ( ) ( 6. nc/m ) ( 8N m/c.5cm 4.5cm 8.988 N m /C ) 9 < < xpess fo 4.5 cm < < 6.5 cm: nd R ( 4.5cm < < 6.5cm) Letting epesent the chge density on the oute sufce, expess fo > 6.5 cm: Substitute in eqution () to obtin: A π R L whee R 6.5 cm. ( π R L) k R R ( > R ) L In (b) we show tht. nc/m. Substitute numeicl vlues to obtin: R ( ) (.nc/m )( 6.5cm).5cm > 6 56N m/c ( 8.854 C / N m )
4 Chpte (b) The sufce chge densities on the nd the outside sufces of the oute conducto e given by: λ πr nd λ outside πr outside Substitute numeicl vlues nd evlute nd outside : 6. nc/m π (.45 m). nc/m nd 6. nc/m outside π (.65 m) 4.7 nc/m. nc/m 54 An infinitely long non-conducting solid cylinde of dius hs nonunifom volume chge density. This density vies linely with R, the pependicul distnce fom its xis, ccoding to ρ(r) βr, whee β is constnt. () Show tht the line chge density of the cylinde is given by λ πβ /. (b) Find expessions fo the electic field fo R < nd R >. Pictue the Poblem Fom symmety considetions, we cn conclude tht the field tngent to the sufce of the cylinde must vnish. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the infinitely long nonconducting cylinde. () Apply Guss s lw to cylindicl sufce of dius nd length L tht is concentic with the infinitely long nonconducting cylinde: n da S o πl n R () πl whee we ve neglected the end es becuse thee is no flux though them. xpess d fo ρ() : d ρ( ) dv ( πl) d π Ld
The lectic Field II: Continuous Chge Distibutions 4 Integte to obtin: d fom to R R πl πl R d πl R Divide both sides of this eqution by L to obtin n expession fo the chge pe unit length λ of the cylinde: L λ πr (b) Substitute fo () nd simplify to obtin: in eqution R πl π L ( < R) Fo > R: πl R Substitute fo in eqution () nd simplify to obtin: πl R ( ) R > R πl R 55 [SSM] An infinitely long non-conducting solid cylinde of dius hs non-unifom volume chge density. This density vies with R, the pependicul distnce fom its xis, ccoding to ρ(r) br, whee b is constnt. () Show tht the line chge density of the cylinde is given by λ πb 4 /. (b) Find expessions fo the electic field fo R < nd R >. Pictue the Poblem Fom symmety; the field tngent to the sufce of the cylinde must vnish. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the infinitely long nonconducting cylinde. () Apply Guss s lw to cylindicl sufce of dius nd length L tht is concentic with the infinitely long nonconducting cylinde: n da S o πl n R () πl whee we ve neglected the end es becuse thee is no flux though them.
44 Chpte xpess fo ρ() b d : d ρ( ) dv b ( πl) πb Ld d Integte to obtin: d fom to R R πbl πbl R 4 4 d πbl 4 R Divide both sides of this eqution by L to obtin n expession fo the chge pe unit length λ of the cylinde: L λ πbr 4 (b) Substitute fo () nd simplify to obtin: in eqution R πbl 4 b π L 4 ( < R) Fo > R: πbl 4 R Substitute fo in eqution () nd simplify to obtin: π bl 4 R ( ) R > R π L 4 br 4 56 An infinitely long, non-conducting cylindicl shell of inne dius nd oute dius hs unifom volume chge density ρ. Find expessions fo the electic field eveywhee. Pictue the Poblem Fom symmety; the field tngent to the sufce of the cylinde must vnish. We cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to find the electic field s function of the distnce fom the centeline of the infinitely long nonconducting cylindicl shell. Apply Guss s lw to cylindicl sufce of dius nd length L tht is concentic with the infinitely long nonconducting cylindicl shell: n da S o πl n R π L whee we ve neglected the end es becuse no flux cosses them.
Fo < R, : The lectic Field II: Continuous Chge Distibutions 45 ( < R ) R xpess fo R < < R : Substitute fo obtin: xpess Substitute fo obtin: fo > R: nd simplify to nd simplify to R R ρv ρπ L ρπ L ρπl ( R ) ρπl ( ) ( R ) R < < R ρπl ( R R ) π L ( R ) ρ ρv ρπb L ρπ L ρπl ( ) ( R R ) > b π L ( R R ) ρ 57 [SSM] The inne cylinde of Figue -4 is mde of nonconducting mteil nd hs volume chge distibution given by ρ(r) C/R, whee C nc/m. The oute cylinde is metllic, nd both cylindes e infinitely long. () Find the chge pe unit length (tht is, the line chge density) on the inne cylinde. (b) Clculte the electic field fo ll vlues of R. Pictue the Poblem We cn integte the density function ove the dius of the inne cylinde to find the chge on it nd then clculte the line chge density fom its definition. To find the electic field fo ll vlues of we cn constuct Gussin sufce in the shpe of cylinde of dius nd length L nd pply Guss s lw to ech egion of the cble to find the electic field s function of the distnce fom its centeline. () Find the chge inne cylinde: inne on the inne R ρ () πcl dv R R C πld d πclr Relte this chge to the line chge density: inne πclr λinne πcr L L
46 Chpte Substitute numeicl vlues nd evlute λ inne : λ inne π ( nc/m)(.5m) 8.8nC/m (b) Apply Guss s lw to cylindicl sufce of dius nd length L tht is concentic with the infinitely long nonconducting cylinde: Substitute to obtin, fo <.5 cm: Substitute numeicl vlues nd evlute R ( <.5 cm): n da S o πl n R πl whee we ve neglected the end es becuse thee is no flux though them. πcl C R ( <.5cm) π L R (.5cm) nc/m 8.854 C /N m <.6kN/C xpess fo.5 cm < < 4.5 cm: Substitute to obtin, fo.5 cm < < 4.5 cm: πclr R (.5cm < < 4.5cm) whee R.5 cm. CπRL π L CR Substitute numeicl vlues nd evlute n (.5 cm < < 4.5 cm): ( ) ( nc/m )(.5m).5cm < 4.5cm R < 9N m/c ( 8.854 C /N m ) Becuse the oute cylindicl shell is conducto: R ( 4.5cm < < 6.5cm) Fo > 6.5 cm, nd: πclr R ( > 6.5 cm) 9 N m/c
The lectic Field II: Continuous Chge Distibutions 47 lectic Chge nd Field t Conducto Sufces 58 An unchged penny is in egion tht hs unifom electic field of mgnitude.6 kn/c diected pependicul to its fces. () Find the chge density on ech fce of the penny, ssuming the fces e plnes. (b) If the dius of the penny is. cm, find the totl chge on one fce. Pictue the Poblem Becuse the penny is in n extenl electic field, it will hve chges of opposite signs induced on its fces. The induced chge is elted to the electic field by /. Once we know, we cn use the definition of sufce chge density to find the totl chge on one fce of the penny. () Relte the electic field to the chge density on ech fce of the penny: Substitute numeicl vlues nd evlute : ( 8.854 C /N m )(.6kN/C) 4.7 nc/m 4. nc/m (b) Use the definition of sufce chge density to obtin: Substitute numeicl vlues nd evlute : π A π π ( 4.7 nc/m )(. m) 4.45pC 59 A thin metl slb hs net chge of zeo nd hs sque fces tht hve -cm-long sides. It is in egion tht hs unifom electic field tht is pependicul to its fces. The totl chge induced on one of the fces is. nc. Wht is the mgnitude of the electic field? Pictue the Poblem Becuse the metl slb is in n extenl electic field, it will hve chges of opposite signs induced on its fces. The induced chge is elted to the electic field by. / Relte the mgnitude of the electic field to the chge density on the metl slb: Use its definition to expess : A L
48 Chpte Substitute fo to obtin: L Substitute numeicl vlues nd evlute : (.m) ( 8.854 C /N m ) 9.4 kn/c.nc 6 A chge of -6. nc is unifomly distibuted on thin sque sheet of non-conducting mteil of edge length. cm. () Wht is the sufce chge density of the sheet? (b) Wht e the mgnitude nd diection of the electic field next to the sheet nd poximte to the cente of the sheet? Pictue the Poblem We cn pply its definition to find the sufce chge density of the nonconducting mteil nd clculte the electic field t eithe of its sufces fom /( ). () Use its definition to find : A 6. nc 5nC/m (. m) (b) The mgnitude of the electic field just outside the sufce of the sheet on the side tht is chged is given by: 8.47 kn/c 5 nc/m 8.854 ( C /N m ) The diection of the field on the side of the sheet tht is chged is the diection of the electic foce cting on test chge. Becuse the sufce is negtively chged, this foce nd, hence, the electic field, is diected towd the sufce. Becuse the sheet is constucted fom non-conducting mteil, no chge is induced on the second sufce of the sheet nd thee is, theefoe, no electic field just outside the sheet sufce on this side. 6 A conducting spheicl shell tht hs zeo net chge hs n inne dius R nd n oute dius R. A positive point chge q is plced t the cente of the shell. () Use Guss s lw nd the popeties of conductos in electosttic equilibium to find the electic field in the thee egions: < R, R < < R, nd > R, whee is the distnce fom the cente. (b) Dw the electic field lines in ll thee egions. (c) Find the chge density on the inne sufce ( R) nd on the oute sufce ( R ) of the shell. Pictue the Poblem We cn constuct Gussin sufce in the shpe of sphee of dius with the sme cente s the shell nd pply Guss s lw to find
The lectic Field II: Continuous Chge Distibutions 49 the electic field s function of the distnce fom this point. The inne nd oute sufces of the shell will hve chges induced on them by the chge q t the cente of the shell. () Apply Guss s lw to spheicl sufce of dius tht is concentic with the point chge: S n da 4π Solving fo yields: () 4π Fo < R, q. Substitute in eqution () nd simplify to obtin: Becuse the spheicl shell is conducto, chge q will be induced on its inne sufce. Hence, fo R < < R : Fo > R, q. Substitute in eqution () nd simplify to obtin: q kq ( < R ) 4π nd R < < R ( ) q kq ( > R ) 4π (b) The electic field lines e shown in the digm to the ight: (c) A chge q is induced on the inne sufce. Use the definition of sufce chge density to obtin: inne q 4πR A chge q is induced on the oute sufce. Use the definition of sufce chge density to obtin: oute q 4πR