Chapter 22 The Electric Field II: Continuous Charge Distributions

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 x y x y + + + + + + + + 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

5 Chpte 6 The electic field just bove the sufce of th hs been mesued to typiclly be 5 N/C pointing downwd. () Wht is the sign of the net chge on th s sufce unde typicl conditions? (b)wht is the totl chge on th s sufce implied by this mesuement? Pictue the Poblem We cn constuct spheicl Gussin sufce t the sufce of th (we ll ssume th is sphee) nd pply Guss s lw to elte the electic field to its totl chge. () Becuse the diection of n electic field is the diection of the foce cting on positively chged object, the net chge on th s sufce must be negtive. (b)apply Guss s lw to spheicl sufce of dius R tht is concentic with th: S n da 4πR n Solve fo th to obtin: th 4 π Rn R k n Substitute numeicl vlues nd evlute : th th 6 ( 6.7 m) ( 5 N/C) 8.988 677 kc 9 N m /C 6 [SSM] A positive point chge of.5 μc is t the cente of conducting spheicl shell tht hs net chge of zeo, n inne dius equl to 6 cm, nd n oute dius equl to 9 cm. () Find the chge densities on the inne nd oute sufces of the shell nd the totl chge on ech sufce. (b) Find the electic field eveywhee. (c) Repet Pt () nd Pt (b) with net chge of +.5 μc plced on the shell. Pictue the Poblem Let the inne nd oute dii of the unchged spheicl conducting shell be R nd R nd q epesent the positive point chge t the cente of the shell. The positive point chge t the cente will induce negtive chge on the inne sufce of the shell nd, becuse the shell is unchged, n equl positive chge will be induced on its oute sufce. To solve Pt (b), 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 electic field s function of the distnce fom this point. In Pt (c) we cn use simil sttegy with the dditionl chge plced on the shell. () xpess the chge density on the inne sufce: q inne inne A

The lectic Field II: Continuous Chge Distibutions 5 xpess the eltionship between the positive point chge q nd the chge induced on the inne sufce q inne : q + q q q inne inne Substitute fo q inne nd A to obtin: inne q 4πR Substitute numeicl vlues nd.5μc π inne evlute inne : 4 (.6m).55 μc/m xpess the chge density on the oute sufce: Becuse the spheicl shell is unchged: q oute oute A q oute + qinne Substitute fo q oute to obtin: oute q 4πR inne Substitute numeicl vlues nd.5μc π oute evlute oute : 4 (.9m).5 μc/m (b) Apply Guss s lw to spheicl sufce of dius tht is concentic with the point chge: Solve fo : S n da 4π () 4π Fo < R 6 cm, q. Substitute in eqution () nd evlute ( < 6 cm) to obtin: 9 q kq ( ) ( 8.988 N m /C )(.5μC) < 6 cm 4π 4 (. N m /C)

5 Chpte Becuse the spheicl shell is conducto, chge q will be induced on its inne sufce. Hence, fo 6 cm < < 9 cm: nd 6 cm < < 9cm ( ) Fo > 9 cm, the net chge the Gussin sufce is q nd: kq 4 ( > 9cm) (. N m /C) (c) Becuse in the conducto: q inne nd inne.5μc.55 μc/m s befoe. xpess the eltionship between the chges on the inne nd oute sufces of the spheicl shell: q oute nd q oute + q inne.5μc.5μ C - q 6. μc inne oute is now given by: oute 6.μC 4 π (.9m).59 μc/m Fo < R 6 cm, q nd ( < 6 cm) is s it ws in (): Becuse the spheicl shell is conducto, chge q will be induced on its inne sufce. Hence, fo 6 cm < < 9 cm: ( < 6 cm) (. N m /C) nd 6 cm < < 9cm ( ) 4 Fo >.9 m, the net chge the Gussin sufce is 6. μc nd: kq 9 4 ( > 9cm) ( 8.988 N m /C )( 6. μ C) ( 5.4 N m /C) 64 If the mgnitude of n electic field in i is s get s. 6 N/C, the i becomes ionized nd begins to conduct electicity. This phenomenon is clled dielectic bekdown. A chge of 8 μc is to be plced on conducting

The lectic Field II: Continuous Chge Distibutions 5 sphee. Wht is the minimum dius of sphee tht cn hold this chge without bekdown? Pictue the Poblem Fom Guss s lw we know tht the electic field t the sufce of the chged sphee is given by k R whee is the chge on the sphee nd R is its dius. The minimum dius fo dielectic bekdown coesponds to the mximum electic field t the sufce of the sphee. Use Guss s lw to expess the electic field t the sufce of the chged sphee: k R xpess the eltionship between k nd R fo dielectic bekdown: mx R min R min k mx Substitute numeicl vlues nd evlute R min : R min 9 ( 8.988 N m /C )( 8μC). 6 N/C cm 65 [SSM] A sque conducting slb cies net chge of 8 μc. The dimensions of the slb e. cm 5. m 5. m. To the left of the slb is n infinite non-conducting flt sheet with chge density. μc/m. The fces of the slb e pllel to the sheet. () Find the chge on ech fce of the slb. (Assume tht on ech fce of the slb the sufce chge is unifomly distibuted, nd tht the mount of chge on the edges of the slb is negligible.) (b) Find the electic field just to the left of the slb nd just to the ight of the slb. Pictue the Poblem () We cn use the fct tht the net chge on the conducting slb is the sum of the chges left nd ight on its left nd ight sufces to obtin line eqution elting these chges. Becuse the electic field is zeo the slb, we cn obtin second line eqution in left nd ight tht we cn solve simultneously with the fist eqution to find left nd ight. (b) We cn find the electic field on ech side of the slb by dding the fields due to the chges on the sufces of the slb nd the field due to the plne of chge.. μ C/m Infinite non-conducting flt sheet left left sheet ight ight

54 Chpte () The net chge on the conducting slb is the sum of the chges on the sufces to the left nd to the ight: Becuse the electic field is equl to zeo the slb: Letting A epesent the e of the chged sufces of the slb nd substituting fo left nd ightt yields: + 8 μc () left ight left ight + sheet left ight + A A sheet Simplifying to obtin: A + sheet left ight Substituting numeicl vlues yields: 5 μc + left ight m o 5μC () (. m). left ight Solve equtions () nd () simultneously to obtin: (b) xpess the totl field just to the left of the slb: 5 μc nd 65 μc left + ight + left of sheet left ight the slb sheet left ight ˆ ˆ ˆ whee ˆ is unit vecto pointing wy fom the slb. Substituting fo left nd ight nd simplifying yields: left the slb of sheet left ight ˆ ˆ ˆ A A A sheet ( left + ight ) ˆ A Substitute numeicl vlues nd evlute : left the slb of left of the slb ( 5. m) (. μc/m ) ( 5 μc + 65μC ) ( 5. m) ( 8.854 C /N m ) ˆ ( 68kN/C)ˆ

The lectic Field II: Continuous Chge Distibutions 55 xpess the totl field just to the ight of sheet + + left ight sufce the slb of the slb ight of the slb: sheet left ight ˆ + ˆ + ˆ Substituting fo left nd ight nd simplifying yields: ight of the slb sheet left ight ˆ + ˆ + ˆ A A A sheet + ( ight + left ) ˆ A Substitute numeicl vlues nd evlute : ight of the slb ( 5. m) (. μc/m ) + ( 65 μc + 5μC ) ( 5. m) ( 8.854 C /N m ) ight of the slb ˆ (.9 MN/C)ˆ Genel Poblems 66 Conside the concentic metl sphee nd spheicl shells tht e shown in Figue -4. The innemost is solid sphee tht hs dius R. A spheicl shell suounds the sphee nd hs n inne dius R nd n oute dius R. The sphee nd the shell e both suounded by second spheicl shell tht hs n inne dius R 4 nd n oute dius R 5. None of these thee objects initilly hve net chge. Then, negtive chge is plced on the inne sphee nd positive chge + is plced on the outemost shell. () Afte the chges hve eched equilibium, wht will be the diection of the electic field between the inne sphee nd the middle shell? (b) Wht will be the chge on the inne sufce of the middle shell? (c) Wht will be the chge on the oute sufce of the middle shell? (d) Wht will be the chge on the inne sufce of the outemost shell? (e) Wht will be the chge on the oute sufce of the outemost shell? (f) Plot s function of fo ll vlues of. Detemine the Concept We cn detemine the diection of the electic field between sphees I nd II by imgining test chge plced between the sphees nd detemining the diection of the foce cting on it. We cn detemine the mount nd sign of the chge on ech sphee by elizing tht the chge on given sufce induces chge of the sme mgnitude but opposite sign on the next sufce of lge dius. () The chge plced on sphee III hs no being on the electic field between sphees I nd II. The field in this egion will be in the diection of the foce exeted on test chge plced between the sphees. Becuse the chge t the cente is negtive, the field will point towd the cente.

56 Chpte (b) The chge on sphee I ( ) will induce chge of the sme mgnitude but opposite sign on sphee II: + (c) The induction of chge + on the inne sufce of sphee II will leve its oute sufce with chge of the sme mgnitude but opposite sign: (d) The pesence of chge on the oute sufce of sphee II will induce chge of the sme mgnitude but opposite sign on the inne sufce of sphee III: + (e) The pesence of chge + on the inne sufce of sphee III will leve the oute sufce of sphee III neutl: (f) A gph of s function of is shown to the ight: 67 [SSM] A lge, flt, nonconducting, non-unifomly chged sufce lies in the x plne. At the oigin, the sufce chge density is +. μc/m. A smll distnce wy fom the sufce on the positive x xis, the x component of the electic field is 4.65 5 N/C. Wht is x smll distnce wy fom the sufce on the negtive x xis? Pictue the Poblem Becuse the diffeence between the field just to the ight of the sufce nd the field just to the left of the sufce is the field due x,pos to the nonunifom sufce chge, we cn expess nd. x,pos x,neg x, neg s the diffeence between xpess the electic field just to the left of the oigin in tems of x,pos nd : x,neg x,pos

The lectic Field II: Continuous Chge Distibutions 57 Substitute numeicl vlues nd evlute x,neg : 4 5. μc/m.65 N/C 8.854 C /N m x, neg 5 kn/c 68 An infinitely long line chge tht hs unifom line chge density equl to.5 μc/m lies pllel to the y xis t x. m. A positive point chge tht hs mgnitude equl to. μc is locted t x. m, y. m. Find the electic field t x. m, y.5 m. Pictue the Poblem Let P denote the point of inteest t (. m,.5 m). The electic field t P is the sum of the electic fields due to the infinite line chge nd the point chge. λ.5 μc/m. y, m q.μc (.,.) P(.,.5).... x, m xpess the esultnt electic field t P: λ + q Find the field t P due the infinite line chge: λ kλ ˆ 9 ( N m /C )(.5μC/m) 8.988 4.m iˆ ( 6.74kN/C)iˆ xpess the field t P due the point chge: Refeing to the digm bove, detemine nd ˆ : q kq ˆ.8m nd ˆ.8944ˆ i.447 ˆj

58 Chpte ( ) Substitute nd evlute q. m,.5 m : q (.8944ˆ i.447 ˆj ) ( 9.48kN/C)(.8944ˆ i.447 ˆj ) 9 ( ) ( 8.988 N m /C )(. μc).m,.5m (.8m) Substitute nd simplify to obtin: ( 8.6kN/C) iˆ ( 4.8 kn/c)j ˆ (. m,.5m) ( 6.74kN/C) iˆ + ( 8.6kN/C) iˆ ( 4.8 kn/c) (.6 kn/c) iˆ ( 4.8kN/C)j ˆ 69 [SSM] A thin, non-conducting, unifomly chged spheicl shell of dius R (Figue -44) hs totl positive chge of. A smll cicul plug is emoved fom the sufce. () Wht is the mgnitude nd diection of the electic field t the cente of the hole? (b) The plug is now put bck in the hole (Figue - 44b). Using the esult of Pt (), find the electic foce cting on the plug. (c) Using the mgnitude of the foce, clculte the electosttic pessue (foce/unit e) tht tends to expnd the sphee. Pictue the Poblem If the ptch is smll enough, the field t the cente of the ptch comes fom two contibutions. We cn view the field in the hole s the sum of the field fom unifom spheicl shell of chge plus the field due to smll ptch with sufce chge density equl but opposite to tht of the ptch cut out. ˆj () xpess the mgnitude of the electic field t the cente of the hole: Apply Guss s lw to spheicl gussin sufce just outside the given sphee: spheicl + shell ( π ) hole enclosed spheicl 4 shell Solve fo spheicl shell to obtin: spheicl shell 4π The electic field due to the smll hole (smll enough so tht we cn tet it s plne sufce) is: hole

The lectic Field II: Continuous Chge Distibutions 59 Substitute nd simplify to obtin: (b) xpess the foce on the ptch: 4π 4π + ( 4π ) dilly outwd 8π F q whee q is the chge on the ptch. Assuming tht the ptch hs dius, expess the popotion between its chge nd tht of the spheicl shell: q π o q 4π 4 Substitute fo q nd in the expession fo F to obtin: F 4 8π π 4 dilly outwd (c) The pessue is the foce exeted on the ptch divided by the e of 4 π the ptch: P 4 π π 7 An infinite thin sheet in the y plne hs unifom sufce chge density +65 nc/m. A second infinite thin sheet hs unifom chge density +45 nc/m nd intesects the y plne t the z xis nd mkes n ngle of º with the xz plne, s shown in Figue -45. Find the electic field t () x 6. m, y. m nd (b) x 6. m, y 5. m. Pictue the Poblem Let the numel efe to the plne with chge density nd the numel to the plne with chge density. We cn find the electic field t the two points of inteest by dding the electic fields due to the chge distibutions of the two infinite plnes. xpess the electic field t ny point in spce due to the chge distibutions on the two plnes: + ()

6 Chpte () xpess the electic field t (6. m,. m) due to plne : 6. ˆ + 65nC/m j 8.85 ( m,.m) ( C /N m ) xpess the electic field t (6. m,. m) due to plne : 6. + 45nC/m ˆ 8.85 ( m,.m) ( C /N m ) ˆj ˆ (.67 kn/c)j ˆ (.54kN/C)ˆ whee ˆ is unit vecto pointing fom plne towd the point whose coodintes e (6. m,. m). Refe to the digm below to obtin: ˆ sin iˆ cos ˆj Substitute to obtin: Substitute in eqution () to obtin: ( 6.m,.m) (.54 kn/c)( sin iˆ cos ˆj ) (.7 kn/c) iˆ + (. kn/c) ˆj ( 6. m,.m) (.67 kn/c) ˆj + (.7 kn/c) iˆ + (. kn/c) (.kn/c) iˆ + (.5kN/C)j ˆ 6. so tht: (b) Note tht ( m,5.m) ( 6.m,.m) ˆ 65nC/m j 8.85 ( 6 m,5m) ˆj ( C / N m ) 6. so tht: Note lso tht ( m,5.m) ( 6.m,.m) ( 6. m,5.m) (.7 kn/c) iˆ (. kn/c)j ˆ + ˆj (.67 kn/c)j ˆ

The lectic Field II: Continuous Chge Distibutions 6 Substitute in eqution () to obtin: ( 6. m,5.m) (.67 kn/c) ˆj + (.7 kn/c) iˆ + (. kn/c) (.kn/c) iˆ + ( 5.9 kn/c)j ˆ 7 Two identicl sque pllel metl pltes ech hve n e of 5 cm. They e septed by.5 cm. They e both initilly unchged. Now chge of +.5 nc is tnsfeed fom the plte on the left to the plte on the ight nd the chges then estblish electosttic equilibium. (Neglect edge effects.) () Wht is the electic field between the pltes t distnce of.5 cm fom the plte on the ight? (b) Wht is the electic field between the pltes distnce of. cm fom the plte on the left? (c) Wht is the electic field just to the left of the plte on the left? (d) Wht is the electic field just to the ight of the plte to the ight? Pictue the Poblem The tnsfe of chge fom the plte on the left to the plte on the ight leves the pltes with equl but opposite chges. Becuse the metl pltes e conductos, the chge on ech plte is completely on the sufce fcing the othe plte. The symbols fo the fou sufce chge densities e shown in the figue. The x component of the electic field due to the chge on sufce L is t points to the left of sufce L nd is L + L t points to the ight of sufce L, whee the +x diection is to the ight. Simil expessions descibe the electic fields due to the othe thee sufce chges. We cn use supeposition of electic fields to find the electic field in ech of the thee egions. ˆj I II III L R L R Define nd so tht: + nd + L R L R

6 Chpte () nd (b) In the egion between the pltes (egion II): x, II L R L R + + Let. Then: Substituting fo nd x, II using the definition of yields: Α Substitute numeicl vlues nd evlute x, II : 8.854.5 nc C N m x, II -6 ( 5 m ) 9 kn/c towd the left (c) The electic field stength just to the left of the plte on the left (egion I) is given by: x, I L R L R (d) The electic field stength just to the ight of the plte on the ight (egion III) is given by: x, III L R L R + + + + + + 7 Two infinite nonconducting unifomly chged plnes lie pllel to ech othe nd to the yz plne. One is t x. m nd hs sufce chge density of.5 μc/m. The othe is t x. m nd hs sufce chge density of 6. μc/m. Find the electic field in the egion () x <. m, (b). m < x <. m, nd (c) x >. m. Pictue the Poblem Let the numel efe to the infinite plne t x. m nd the numel to the plne t x. m nd let I, II, nd III identify the egions to the left of plne, between the plnes, nd to the ight of plne, espectively. We cn use the expession fo the electic field of in infinite plne of chge to expess the electic field due to ech plne of chge in ech of the thee

The lectic Field II: Continuous Chge Distibutions 6 egions. Thei sum will be the esultnt electic field in ech egion..5μc/m 6.μC/m I II III.. x, m The esultnt electic field is the sum of the fields due to plnes nd : () The field due to plne in egion I is given by: The field due to plne in egion I is given by: Substitute fo nd in eqution () nd simplify to obtin: + () ( iˆ ) ( iˆ ) ( iˆ ) ( iˆ + ) iˆ + Substitute numeicl vlues nd evlute : ( 8.854 C /N m ).5μC/m ( 4kN/C)iˆ + 6.μC/m iˆ (b) The field due to plne in egion II is given by: The field due to plne in egion II is given by: Substitute fo nd in eqution () nd simplify to obtin: iˆ ( iˆ ) ( iˆ ) ( iˆ + ) iˆ

64 Chpte Substitute numeicl vlues nd evlute : ( 8.854 C /N m ).5μC/m ( 56kN/C)iˆ 6.μC/m iˆ (c) The field due to plne in egion III is given by: The field due to plne in egion III is given by: Substitute fo nd in eqution () nd simplify to obtin: iˆ iˆ ( iˆ ) ( iˆ + ) iˆ + Substitute numeicl vlues nd evlute : ( 8.854 C /N m ).5μC/m ( 4kN/C)iˆ + 6.μC/m 7 [SSM] A quntum-mechnicl tetment of the hydogen tom shows tht the electon in the tom cn be teted s smeed-out distibution of negtive chge of the fom ρ() ρ e /. Hee epesents the distnce fom the cente of the nucleus nd epesents the fist Boh dius which hs numeicl vlue of.59 nm. Recll tht the nucleus of hydogen tom consists of just one poton nd tet this poton s positive point chge. () Clculte ρ, using the fct tht the tom is neutl. (b) Clculte the electic field t ny distnce fom the nucleus. Pictue the Poblem Becuse the tom is unchged, we know tht the integl of the electon s chge distibution ove ll of spce must equl its chge q e. vlution of this integl will led to n expession fo ρ. In (b) we cn expess the esultnt electic field t ny point s the sum of the electic fields due to the poton nd the electon cloud. iˆ () Becuse the tom is unchged, the integl of the electon s chge distibution ove ll of spce must equl its chge e: Substitute fo ρ() nd simplify to obtin: e ρ e ρ e 4πρ () dv ρ() e 4π d d 4π d

The lectic Field II: Continuous Chge Distibutions 65 Use integl tbles o integtion by pts to obtin: 4 d e Substitute fo d e to obtin: 4 4 ρ π πρ e Solving fo ρ yields: e π ρ (b) The field will be the sum of the field due to the poton nd tht of the electon chge cloud: p cloud + xpess the field due to the electon cloud: () ( ) cloud k whee () is the net negtive chge enclosed distnce fom the poton. Substitute fo nd to obtin: p cloud () ( ) k ke + () () is given by: ' ' 4 ' ') ( ' 4 ) ( ' d e d ρ π ρ π Fom Pt (), e π ρ : ' ' 4 ' ' 4 ) ( ' ' d e e d e e π π Fom tble of integls: ( ) [ ] ( ) ( ) + 4 4 4 e e e e e e dx e x x

66 Chpte Substituting fo ' ' ' d e in the expession fo nd simplifying yields: () ( ) + 4 ) ( e e e Substitute fo in eqution () nd simplify to obtin: () () ( ) ( ) + + 4 4 e e ke e e ke ke 74 A unifomly chged ing hs dius, lies in hoizontl plne, nd hs negtive chge given by. A smll pticle of mss m hs positive chge given by q. The smll pticle is locted on the xis of the ing. () Wht is the minimum vlue of q/m such tht the pticle will be in equilibium unde the ction of gvity nd the electosttic foce? (b) If q/m is twice the vlue clculted in Pt (), whee will the pticle be when it is in equilibium? xpess you nswe in tems of Pictue the Poblem We cn pply the condition fo tnsltionl equilibium to the pticle nd use the expession fo the electic field on the xis of ing chge to obtin n expession fo q/m. Doing so will led us to the conclusion tht q/m will be minimum when z is mximum nd tht z mximizes z. m,q y z ing

The lectic Field II: Continuous Chge Distibutions 67 () Apply F z to the pticle: q g q z mg () m z Note tht this esult tells us tht the minimum vlue of q/m will be whee the field due to the ing is getest. xpess the electic field long the z kz z z + xis due to the ing of chge: ( ) Diffeentite this expession with espect to z to obtin: d x dz d k dz k d ( z ) z ( z ) z + + dx k ( z + ) ( z + ) ( z + ) z()( ) z + ( z) ( z + ) z ( z + ) k ( z + ) ( z + ) Set this expession equl to zeo fo extem nd simplify: Solve fo z to obtin: ( z + ) z ( z + ) ( z + ) ( + ) z ( z + ), z, nd z + z z ± s cndidtes fo mxim o minim. You cn eithe plot gph of z o evlute its second deivtive t these points to show tht it is mximum t: z Substitute to obtin n expession z,mx : k z,mx + k 7

68 Chpte Substitute in eqution () to obtin: q m 7g k (b) If q/m is twice s get s in (), then the electic field should be hlf its vlue in (), tht is: k kz ( z + ) 7 o 7 4 z 6 z + Let u z / nd simplify to obtin: u + u 4u + The following gph of pogm. f ( u) u + u 4u + ws plotted using spedsheet 5 5 f () -5 - -5 - -5 -..5..5..5..5 4. Use you gphing clculto o tilnd-eo methods to obtin: The coesponding z vlues e: u.489 nd u. 596 z. 5 nd z. 9 The condition fo stble equilibium position is tht the pticle, when displced fom its equilibium position, expeiences estoing foce; tht is, foce tht cts towd the equilibium position. When the pticle in this poblem is just bove its equilibium position the net foce on it must be downwd nd when it is just below the equilibium position the net foce on it must be upwd. Note tht the electic foce is zeo t the oigin, so the net foce thee is downwd nd emins downwd to the fist equilibium position s the weight foce exceeds the electic foce in this intevl. The net foce is upwd between the fist nd

The lectic Field II: Continuous Chge Distibutions 69 second equilibium positions s the electic foce exceeds the weight foce. The net foce is downwd below the second equilibium position s the weight foce exceeds the electic foce. Thus, the fist (highe) equilibium position is stble nd the second (lowe) equilibium position is unstble. You might lso find it instuctive to use you gphing clculto to plot gph of the electic foce (the gvittionl foce is constnt nd only shifts the gph of the totl foce downwd). Doing so will poduce gph simil to the one shown in the sketch to the ight..9.5 F fce z Note tht the slope of the gph is negtive on both sides of.5 whees it is positive on both sides of.9; futhe evidence tht.5 is position of stble equilibium nd.9 position of unstble equilibium. 75 A long, thin, non-conducting plstic od is bent into cicul loop tht hs dius. Between the ends of the od shot gp of length l, whee l <<, emins. A positive chge of mgnitude is evenly distibuted on the loop. () Wht is the diection of the electic field t the cente of the loop? xplin you nswe. (b) Wht is the mgnitude of the electic field t the cente of the loop? Pictue the Poblem The loop with the smll gp is equivlent to closed loop nd chge of l ( πr ) t the gp. The field t the cente of closed loop of unifom line chge is zeo. Thus the field is entiely due to the chge l πr. ( ) () xpess the field t the cente of the loop: cente + () loop gp Relte the field t the cente of the loop to the chge in the gp: gp kq ˆ Use the definition of line chge q l λ q l π π

7 Chpte density to elte the chge in the gp to the length of the gp: Substitute fo q to obtin: gp kl π ˆ Substituting in eqution () yields: cente kl π kl ˆ π ˆ If is positive, the field t the oigin points dilly outwd towd the gp. (b) Fom ou esult in () we see tht the mgnitude of kl cente is: π cente 76 A non-conducting solid sphee tht is. m in dimete nd hs its cente on the x xis t x 4. m hs unifom volume chge of density of +5. μc/m. Concentic with the sphee is thin non-conducting spheicl shell tht hs dimete of.4 m nd unifom sufce chge density of.5 μc/m. Clculte the mgnitude nd diection of the electic field t () x 4.5 m, y, (b) x 4. m, y. m, nd (c) x. m, y. m. Pictue the Poblem We cn find the electic fields t the thee points of inteest, lbeled,, nd in the digm, by dding the electic fields due to the chge distibutions on the nonconducting sphee nd the spheicl shell. y, m...5μc/m..6 m 4. x, m. m ρ 5. μc/m xpess the electic field due to the sphee + () shell

nonconducting sphee nd the spheicl shell t ny point in spce: The lectic Field II: Continuous Chge Distibutions 7 () Becuse (4.5 m, ) is the spheicl shell: Apply Guss s lw to spheicl sufce the nonconducting sphee to obtin: shell ( 4.5 m,) sphee 4π () kρiˆ vlute (.5 m) : sphee sphee 4π ( 94.kN/C)iˆ 9 (.5m) ( 8.988 N m /C )( 5. μc/m )(.5m) iˆ Substitute in eqution () to obtin: Find the mgnitude nd diection 4.5 m, : of ( ) ( 4.5 m,) ( 94.kN/C) iˆ + ( 94.kN/C)iˆ ( 4.5 m,) 94kN/C nd θ (b) Becuse (4. m,. m) is the spheicl shell: shell ( 4. m,. m) vlute (. m) : sphee 9 4π ( ) ( 8.988 N m /C )( 5. μc/m )(.6 m).m sphee (.6 kn/c)j ˆ ( ).m ˆj ˆj + (.6 kn/c)j ˆ Substitute in eqution () to obtin: ( 4. m,. m) (.6 kn/c)

7 Chpte Find the mgnitude nd diection 4. m,. m : of ( ) (c) Becuse (. m,. m) outside the spheicl shell: nd ( 4.m,.m).6kN/C θ 9 kshell () ˆ shell whee ˆ is unit vecto pointing fom (4. m, ) to (. m,. m). vlute shell : shell A 4π shell (.5 μc/m )(.m) 7.4 μc Refe to the following digm to find ˆ nd : y, m (.,.)..66 m nd ˆ.5547ˆ i +.8ˆj. 4. x, m Substitute numeicl vlues nd evlute (. m,. m) shell : shell 9 ( ) ( 8.988 N m /C )( 7.4 μc).66 m (.66 m) ( 8.77 kn/c)(.5547iˆ +.8ˆj ) (.4kN/C) iˆ + ( 5.6kN/C)j ˆ ˆ xpess the electic field due to the chged nonconducting sphee t distnce fom its cente tht is gete thn its dius: sphee k sphee () ˆ

The lectic Field II: Continuous Chge Distibutions 7 Find the chge on the sphee: vlute (.6m) : sphee sphee ρv sphee ( 5. μc/m )(.6 m) π 4.54 μc 4 sphee 9 ( ) ( 8.988 N m /C )( 4.54 μc).m,.m (.66 m) Substitute in eqution () to obtin: (.8 kn/c)(.5547iˆ +.8ˆj ) (.75 kn/c) iˆ + (.6 kn/c)j ˆ ˆ (.8 kn/c) (. m,. m) (.4kN/C) iˆ + ( 5.6kN/C) ˆj + (.75kN/C) + (.6 kn/c) ˆj ( 8.675kN/C) iˆ + (.kn/c)j ˆ Find the mgnitude nd diection of (. m,. m) : iˆ ˆ nd (. m,. m) ( 8.675 kn/c) + (.kn/c) 5.6kN/C.kN/C θ tn 4 8.675kN/C 77 An infinite non-conducting plne sheet of chge tht hs sufce chge density +. μc/m lies in the y.6 m plne. A second infinite non-conducting plne sheet of chge tht hs sufce chge density of. μc/m lies in the x. m plne. Lstly, non-conducting thin spheicl shell of dius of. m nd tht hs its cente in the z plne t the intesection of the two chged plnes hs sufce chge density of. μc/m. Find the mgnitude nd diection of the electic field on the x xis t () x.4 m nd (b) x.5 m. Pictue the Poblem Let the numel efe to the infinite plne whose chge density is nd the numel to the infinite plne whose chge density is. We cn find the electic fields t the two points of inteest by dding the electic fields due to the chge distibutions on the infinite plnes nd the sphee.

74 Chpte y, m. C/m μ.μc/m.4.. 5 x, m.6 (.,.6).μC/m xpess the electic field due to the infinite plnes nd the sphee t ny point in spce: () Becuse (.4 m, ) is the sphee: + + () sphee sphee (.4 m,) Find the field t (.4 m, ) due to plne : ˆ.μC/m j 8.854 ( ) ˆ.4m, j ( C /N m ) Find the field t (.4 m, ) due to plne : (.4m,) ( i ˆ).μC/m 8.854 ˆ Substitute in eqution () to obtin: ( 69.4 kn/c)j ˆ ( ) ( i ) (.9 kn/c C /N m )i (.4 m,) + ( 69.4 kn/c) ˆj + (.9 kn/c) iˆ (.9 kn/c) iˆ + ( 69.4 kn/c)j ˆ Find the mgnitude nd diection of (.4 m,) : nd (.4 m,) (.9 kn/c) + ( 69.4 kn/c).6 kn/c 4kN/C 69.4 kn/c θ tn 56..9 kn/c 56. ˆ

The lectic Field II: Continuous Chge Distibutions 75 (b) Becuse (.5 m, ) is outside the sphee: ksphee (.4m,) ˆ sphee whee ˆ is unit vecto pointing fom (. m,.6 m) to (.5 m, ). vlute sphee : sphee A 4πR 4π sphee (. μc/m )(.m) 7.7μC Refeing to the digm bove, detemine nd ˆ :.66 m nd ˆ.985ˆ i +.74 ˆj Substitute nd evlute (.5 m,) sphee : sphee 9 ( ) ( 8.988 N m /C )( 7.7 μc ).5 m, (.66 m) Find the field t (.5 m, ) due to plne : ( 9.8 kn/c)(.985ˆ i +.74 ˆj ) (.5 kn/c) iˆ + ( 48. kn/c) ˆj ˆ.μC/m j 8.854 ( ) ˆ.5 m, j ( C /N m ) Find the field t (.5 m, ) due to plne :. C/m iˆ μ 8.854 (.5m,) ( C /N m ) Substitute in eqution () to obtin: ˆ ( 69.4 kn/c)j ˆ iˆ (.9 kn/c)iˆ (.4 m,) (.5 kn/c) iˆ + ( 48.9 kn/c) ˆj + ( 69.4 kn/c) ˆj + (.9 kn/c) (.5 kn/c) iˆ + (. kn/c)j ˆ iˆ

76 Chpte Find the mgnitude nd diection of (.5 m,) : nd (.5 m,) (.5kN/C) + (. kn/c) 6kN/C. kn/c θ tn 5.5kN/C 78 An infinite non-conducting plne sheet lies in the x. m plne nd hs unifom sufce chge density of +. μc/m. An infinite non-conducting line chge of unifom line chge density 4. μc/m psses though the oigin t n ngle of 45.º with the x xis in the xy plne. A solid non-conducting sphee of volume chge density 6. μc/m nd dius.8 m is centeed on the x xis t x. m. Clculte the mgnitude nd diection of the electic field in the z plne t x.5 m, y.5 m. Pictue the Poblem Let P epesent the point of inteest t (.5 m,.5 m). We cn find the electic field t P by dding the electic fields due to the infinite plne, the infinite line, nd the sphee. Once we ve expessed the field t P in vecto fom, we cn find its mgnitude nd diection. xpess the electic field t P: plne + line + sphee Find plne t P: plne iˆ. μc/m 8.854 (.9 kn/c)iˆ ( C /N m ) iˆ xpess line t P: line kλ ˆ Refe to the following figue to obtin:.5m iˆ.5m ˆ ( ) ( )j ˆ nd (.77) i (.77)j ˆ ˆ

The lectic Field II: Continuous Chge Distibutions 77.5 y, m 45º 4. λ μ. C/m ' P ρ 6. μc/m.5. μc/m. x, m Substitute nd simplify to obtin: line 9 ( N m /C )( 4.μC/m) (.77 ) iˆ (.77 ) 8.988.77 m (.7 kn/c)(.77) iˆ (.77) [ ˆj ] [ ˆj ] ( 7.9 kn/c) iˆ + ( 7.9 kn/c) ˆj Letting epesent the distnce fom the cente of the sphee to P, pply Guss s lw to spheicl sufce of dius centeed t ( m, ) to obtin n expession fo sphee t P: Refe to the digm used bove to obtin: 4π k' ' ˆ sphee ρ whee ˆ ' is diected towd the cente of the sphee. ' (.5m) iˆ (.5m)j ˆ nd ˆ '.77 ( ) iˆ (.77)j ˆ Substitute numeicl vlues nd simplify to obtin: sphee [ ˆj ] 9 ( 8.988 N m /C )(.77 m)( 6. C/m ) (.77) iˆ + (.77) 4 π μ (.9 kn/c)( iˆ + ˆj ) (.9 kn/c) iˆ + (.9 kn/c)j ˆ vluting yields: (.9 kn/c) iˆ + ( 7.9 kn/c) iˆ + ( 7.9 kn/c) ˆj + (.9 kn/c) + (.9 kn/c) ˆj ( 54. kn/c) iˆ + ( 84.9 kn/c)j ˆ iˆ

78 Chpte Finlly, find the mgnitude nd diection of : ( 54. kn/c) + ( 84.9kN/C) 4kN/C nd 54. kn/c θ tn 84.9 kn/c 79 [SSM] A unifomly chged, infinitely long line of negtive chge hs line chge density of λ nd is locted on the z xis. A smll positively chged pticle tht hs mss m nd chge q is in cicul obit of dius R in the xy plne centeed on the line of chge. () Deive n expession fo the speed of the pticle. (b) Obtin n expession fo the peiod of the pticle s obit. Pictue the Poblem () We cn pply Newton s nd lw to the pticle to expess its speed s function of its mss m, chge q, nd the dius of its pth R, nd the stength of the electic field due to the infinite line chge. (b) The peiod of the pticle s motion is the tio of the cicumfeence of the cicle in which it tvels divided by its obitl speed. x z R λ m,q y () Apply Newton s nd lw to v the pticle to obtin: Fdil q m R whee the inwd diection is positive. Solving fo v yields: The stength of the electic field t distnce R fom the infinite line chge is given by: v qr m kλ R Substitute fo nd simplify to obtin: v kqλ m

The lectic Field II: Continuous Chge Distibutions 79 (b) The speed of the pticle is equl to the cicumfeence of its obit divided by its peiod: πr v T T πr v Substitute fo v nd simplify to obtin: T πr m kqλ 8 A sttiony ing of dius tht lies in the yz plne hs unifomly distibuted positive chge. A smll pticle tht hs mss m nd negtive chge q is locted t the cente of the ing. () Show tht if x <<, the electic field long the xis of the ing is popotionl to x. (b) Find the foce on the pticle of mss m s function of x. (c) Show tht if the pticle is given smll displcement in the +x diection, it will pefom simple hmonic motion. (d) Wht is the fequency of tht motion? Pictue the Poblem Stting with the eqution fo the electic field on the xis of ing chge, we cn fcto the denominto of the expession to show tht, fo x <<, x is popotionl to x. We cn use F x q x to expess the foce cting on the pticle nd pply Newton s nd lw to show tht, fo smll displcements fom equilibium, the pticle will execute simple hmonic motion. Finlly, we cn find the peiod of the motion fom its ngul fequency, which we cn obtin fom the diffeentil eqution of motion. () xpess the electic field on the kx x x + xis of the ing of chge: ( ) Fcto fom the denominto of x to obtin: (b) xpess the foce cting on the pticle s function of its chge nd the electic field: x kx x + kx x + povided x <<. F q kq x x x k x

8 Chpte (c) Becuse the negtively chged pticle expeiences line estoing foce, we know tht its motion will be simple hmonic. Apply Newton s nd lw to the negtively chged pticle to obtin: (d) Relte the fequency of the simple hmonic motion to its ngul fequency: d x kq m x dt o d x kq + x dt m the diffeentil eqution of simple hmonic motion. ω f () π Fom the diffeentil eqution we kq kq ω ω hve: m m Substitute fo ω in eqution () kq nd simplify to obtin: f π m 8 [SSM] The chges nd q of Poblem 8 e +5. μc nd 5. μc, espectively, nd the dius of the ing is 8. cm. When the pticle is given smll displcement in the x diection, it oscilltes bout its equilibium position t fequency of.4 Hz. () Wht is the pticle s mss? (b) Wht is the fequency if the dius of the ing is doubled to 6. cm nd ll othe pmetes emin unchnged? Pictue the Poblem Stting with the eqution fo the electic field on the xis of ing chge, we cn fcto the denominto of the expession to show tht, fo x <<, x is popotionl to x. We cn use F x q x to expess the foce cting on the pticle nd pply Newton s nd lw to show tht, fo smll displcements fom equilibium, the pticle will execute simple hmonic motion. Finlly, we cn find the ngul fequency of the motion fom the diffeentil eqution nd use this expession to find the fequency of the motion when the dius of the ing is doubled nd ll othe pmetes emin unchnged. () xpess the electic field on the kx x x + xis of the ing of chge: ( )

The lectic Field II: Continuous Chge Distibutions 8 Fcto fom the denominto of x to obtin: xpess the foce cting on the pticle s function of its chge nd the electic field: Becuse the negtively chged pticle expeiences line estoing foce, we know tht its motion will be simple hmonic. Apply Newton s nd lw to the negtively chged pticle to obtin: The ngul fequency of the simple hmonic motion of the pticle is given by: kx x x + kx x + povided x <<. F q kq x x x k x d x kq m x dt o d x kq + x dt m the diffeentil eqution of simple hmonic motion. kq ω () m Solving fo m yields: kq m ω kq 4π f Substitute numeicl vlues nd evlute m: 9 N m 8.988 C m 4π ( 5. μc )( 5.μC) (.4 s ) ( 8. cm).997 kg (b) xpess the ngul fequency of the motion if the dius of the ing is doubled: kq ω ' () m ( )

8 Chpte Divide eqution () by eqution () to obtin: ω' πf' m ( ) ω πf kq kq m 8 Solve fo f to obtin: f ' f 8.4 Hz 8.8Hz 8 If the dius of the ing in Poblem 8 is doubled while keeping the line chge density on the ing the sme, does the fequency of oscilltion of the pticle chnge? If so, by wht fcto does it chnge? Pictue the Poblem Stting with the eqution fo the electic field on the xis of ing chge, we cn fcto the denominto of the expession to show tht, fo x <<, x is popotionl to x. We cn use F x q x to expess the foce cting on the pticle nd pply Newton s nd lw to show tht, fo smll displcements fom equilibium, the pticle will execute simple hmonic motion. Finlly, we cn find the ngul fequency of the motion fom the diffeentil eqution nd use this expession to find its vlue when the dius of the ing is doubled while keeping the line chge density on the ing constnt. xpess the electic field on the kx x x + xis of the ing of chge: ( ) Fcto fom the denominto of x to obtin: xpess the foce cting on the pticle s function of its chge nd the electic field: kx x x + kx x + povided x <<. F q kq x x x k x

The lectic Field II: Continuous Chge Distibutions 8 Becuse the negtively chged pticle expeiences line estoing foce, we know tht its motion will be simple hmonic. Apply Newton s nd lw to the negtively chged pticle to obtin: The ngul fequency of the simple hmonic motion of the pticle is given by: xpess the ngul fequency of the motion if the dius of the ing is doubled while keeping the line chge density constnt (tht is, doubling ): d x kq m x dt o d x kq + x, dt m the diffeentil eqution of simple hmonic motion. kq ω () m ( ) ( ) kq ω ' () m Divide eqution () by eqution () to obtin: ω' πf ' ω πf kq m ( ) ( ) kq m Yes. The fequency chnges by fcto of.5. 8 A unifomly chged non-conducting solid sphee of dius R hs its cente t the oigin nd hs volume chge density of ρ. () Show tht t point ρ within the sphee distnce fom the cente ˆ. (b) Mteil is emoved fom the sphee leving spheicl cvity of dius b R/ with its cente t x b on the x xis (Figue -46). Clculte the electic field t points nd shown in Figue -46. Hint: Model the sphee-with-cvity s two unifom sphees of equl positive nd negtive chge densities. Pictue the Poblem In Pt (), you cn pply Guss s lw to expess s function of fo the unifomly chged nonconducting sphee with its cente t the oigin. In Pt (b), you cn use the hint to expess the field t geneic point P(x,y) in the cvity s the sum of the fields due to equl positive nd negtive chge densities nd then evlute this expession t points nd.

84 Chpte () The electic field t distnce fom the cente of the unifomly chged nonconducting sphee is given by: Apply Guss s lw to spheicl sufce of dius centeed t the oigin to obtin: ρ ˆ () whee ˆ is unit vecto pointing dilly outwd. S ( π ) nda ρ 4 Relte to the chge density ρ: ρ 4 π 4 ρπ Substitute fo : Solve fo ρ to obtin: ρ ( 4π ) ρ ρ 4ρπ Substitute fo in eqution () to obtin: ρ ρ ˆ (b) The electic field t point P(x,y) is the sum of the electic fields due to the two chge distibutions: Apply Guss s lw to spheicl sufce of dius centeed t x b R/ to obtin: ˆ ˆ ρ + ρ ρ + ρ ' () whee ˆ ' is unit vecto noml to spheicl Gussin sufce whose cente is t x b. S ( π' ) nda ρ 4 Relte ρ: to the chge density ρ 4 π' 4 ρπ' Substitute fo to obtin: Solving fo ρ yields: Substitute fo ρ nd ρ in eqution () to obtin: ρ ρ ( 4π ' ) ρ 4π ' ρ ' ρ ρ ˆ ' ˆ ()

The vectos given by: ˆ nd The lectic Field II: Continuous Chge Distibutions 85 ' ' ˆ ' e x iˆ + yˆ j nd ' ( x b) iˆ + yˆj whee x nd y e the coodintes of ny point in the cvity. Substitute fo ˆ nd ' ˆ ' in eqution () nd simplify to obtin: ( iˆ ˆ ρ j) [( ) iˆ ˆ ρb x + y x b + yj] iˆ ρ Becuse is independent of x nd y: ρb iˆ 84 Show tht the electic field thoughout the cvity of Poblem 8b is unifom nd is given by ρ bˆ i. Pictue the Poblem The electic field in the cvity is the sum of the electic field due to the unifom nd positive chge distibution of the sphee whose dius is nd the electic field due to ny chge in the spheicl cvity whose dius is b. You cn use the hint given in Poblem 8 to expess the field t geneic point P(x,y) in the cvity s the sum of the fields due to equl positive nd ρ negtive chge densities to show tht biˆ. The electic field t point P(x,y) is the sum of the electic fields due to the two chge distibutions: Apply Guss s lw to spheicl sufce of dius centeed t x b R/ to obtin: ˆ ˆ ρ + ρ ρ + ρ ' () whee ˆ ' is unit vecto noml to spheicl Gussin sufce whose cente is t x b. S ( π' ) nda ρ 4 Relte ρ: to the chge density ρ 4 π ' 4 ρπ ' Substitute fo to obtin: Solving fo ρ yields: ρ ρ ( 4π ' ) ρ 4π ' ρ '

86 Chpte Fom Poblem 8: Substitute fo ρ nd ρ in eqution () to obtin: The vectos ˆ nd ' ' ˆ ' e given by: ρ ρ ρ ρ ˆ ' ˆ () x iˆ + yˆ j nd ' ( x b) iˆ + yˆj whee x nd y e the coodintes of ny point in the cvity. Substitute fo ˆ nd ' ˆ ' in eqution () nd simplify to obtin: ( iˆ ˆ ρ j) [( ) iˆ ˆ ρ x + y x b + yj] biˆ ρ 85 The cvity in Poblem 8b is now filled with unifomly chged nonconducting mteil with totl chge of. Clculte the new vlues of the electic field t points nd shown in Figue -46. Pictue the Poblem The electic field t geneic point P(x,y) in the cvity is the sum of the fields due to the positive chge density nd the totl chge. The electic field t point P(x,y) is the sum of the electic fields due to the two chge distibutions: ρ + ρ + () whee ˆ ' is unit vecto noml to spheicl Gussin sufce whose cente is t x b. Fom Poblem 8: Substituting in eqution () yields: Assuming tht the cvity is filled with positive chge : The vectos ˆ nd ' ' ˆ ' e given by: ρ + ρ ρb iˆ ρb iˆ + ρb iˆ + 4π b 'ˆ x iˆ + yˆ j nd ' ( x b) iˆ + yˆj whee x nd y e the coodintes of ny point in the cvity.

The lectic Field II: Continuous Chge Distibutions 87 Substitute fo ˆ nd ' ˆ ' nd simplify to obtin: At point, x b nd y : ρb ρb iˆ + 4π b ( b,) iˆ ( b b) 4π b [( x b) iˆ + yj ˆ] b [ iˆ ρ ] + iˆ 4π b At point, x nd y : ρb 4π b (,) iˆ + ( b) b [ iˆ ρ ] iˆ 4π b 86 A smll Gussin sufce in the shpe of cube hs fces pllel to the xy, xz, nd yz plnes (Figue -47) nd is in egion in which the electic field is pllel to the x xis. () Using the diffeentil ppoximtion, show tht the net electic flux of the electic field out of the Gussin sufce is given by φ net x x ΔV, whee ΔV is the volume enclosed by the Gussin sufce. (b) Using Guss s lw nd the esults of Pt () show tht x x ρ, whee ρ is the volume chge density the cube. (This eqution is the one-dimensionl vesion of the point fom of Guss s lw.) Pictue the Poblem Let the coodintes of one cone of the cube be (x,y,z), nd ssume tht the sides of the cube e Δx, Δy, nd Δz nd compute the flux though the fces of the cube tht e pllel to the yz plne. The net flux of the electic field out of the gussin sufce is the diffeence between the flux out of the sufce nd the flux into the sufce. () The net flux out of the cube is given by: φ net φ ( x + Δx) φ( x) Use Tylo seies expnsion to expess the net flux though fces of the cube tht e pllel to the yz plne: φ ( x) + ( Δx) φ' ( x) + ( Δx) φ' '( x) +... φ( x) ( Δx) φ' ( x) + ( Δx) φ' '( )... net φ x + Neglecting tems highe thn fist ode we hve: φ Δxφ' net ( x)

88 Chpte Becuse the electic field is in the x diection, φ (x) is: ( x) x ΔyΔz φ nd φ ' x x ( x) ΔyΔz Substitute fo φ (x) to obtin: φ net x Δx x x ΔV x x x ( ΔyΔz) ( ΔxΔyΔz) (b) Fom Guss s lw, the net flux though ny sufce is: qencl ρ φ net ΔV Fom Pt (): φ net x ΔV x qute these two expessions nd simplify to obtin: x x Δ ρ V Δ V x x ρ 87 [SSM] Conside simple but supisingly ccute model fo the hydogen molecule: two positive point chges, ech hving chge +e, e plced unifomly chged sphee of dius R, which hs chge equl to e. The two point chges e plced symmeticlly, equidistnt fom the cente of the sphee (Figue -48). Find the distnce fom the cente,, whee the net foce on eithe point chge is zeo. Pictue the Poblem We cn find the distnce fom the cente whee the net foce on eithe chge is zeo by setting the sum of the foces cting on eithe point chge equl to zeo. ch point chge expeiences two foces; one Coulomb foce of epulsion due to the othe point chge, nd the second due to tht fction of the sphee s chge tht is between the point chge nd the cente of the sphee tht cetes n electic field t the loction of the point chge. Apply F to eithe of the F F Coulomb field () point chges: xpess the Coulomb foce on the poton: The foce exeted by the field is: ke F Coulomb F field e ke ( ) 4

The lectic Field II: Continuous Chge Distibutions 89 Apply Guss s lw to spheicl sufce of dius centeed t the oigin: ( 4π ) enclosed Relte the chge density of the electon sphee to enclosed : e enclosed 4 4 πr π e enclosed R Substitute fo enclosed : Solve fo to obtin: Substitute fo FCoulomb nd Ffield in eqution (): e ( 4π ) R e e F field π R π R ke 4 o ke 4 e π R ke R R R 8 88 An electic dipole tht hs dipole moment of p is locted t pependicul distnce R fom n infinitely long line chge tht hs unifom line chge density λ. Assume tht the dipole moment is in the sme diection s the field of the line of chge. Detemine n expession fo the electic foce on the dipole. Pictue the Poblem We cn find the field due to the infinitely long line chge fom kλ nd the foce tht cts on the dipole using F pd d. xpess the foce cting on the dipole: The electic field t the loction of the dipole is given by: d F p () d kλ Substitute fo in eqution () to d kλ kλp F p obtin: d whee the minus sign indictes tht the dipole is ttcted to the line chge.

9 Chpte