Problem 1. Solution: a) The coordinate of a point on the disc is given by r r cos,sin,0. The potential at P is then given by. r z 2 rcos 2 rsin 2
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1 Prolem Consider disc of chrge density r r nd rdius R tht lies within the xy-plne. The origin of the coordinte systems is locted t the center of the ring. ) Give the potentil t the point P,,z in terms of,r, ndz. ) We next put conducting plne into the z d plne. The potentil of the conducting plne is fixed t V. Compute the totl potentil t point P,,z. c) If the totl chrge, Q, on the disc is fixed, find the chrge density in terms of Q nd use it to otin the form of tot P in terms of Q,R,z in the limit R z,d up to leding order in z/r? d) Give n explicit form of the induced chrge density t P,,d in the limit R d using the results of prt c). Solution: ) The coordinte of point on the disc is given y r rcos,sin, The potentil t P is then given y P 4 dq P r 4 d r r P r R 4 d rdr R 4 d dr r R z r dr r z r R z R z ln R z R z r z rcos rsin where I used dr r z r r z r z ln r z r ) Using the method of imges, the imge chrge is disc of chrge density r r nd rdius R tht is prllel to the conducting plne t distnce of d from the first disc. The resulting totl potentil is then
2 tot P R z R z ln R z R z R d z R d z ln R d z R d z c) If the totl chrge, Q, on the disc is fixed, wht is the form of tot P in terms of Q,R,z in the limit R z,d up to leding order in z/r? Ifthechrgeisfixed,wehve or nd thus Q dq d R rdr r R Q R tot P Q R R z R z ln R z R z Q R R d z R d z ln R d z R d z Q 8 R R z R z ln R z R z Q 8 R R d z R d z ln R d z R d z We first consider the first two terms which yield Q 8 R z R d z R Q z z d 6 R The second set yields Q 6 R 4zd 4d Qd z d 4 R
3 Q 8 R z ln R z R z d z ln Q 8 R z ln R z d z ln R d z This is the leding contriution. R d z R d z d) Give n explicit form of the induced chrge density t P,,d in the limit R d using the results of prt c). The chrge density follows from tot z nd thus tot Q z z zd 8R z ln Q zln z z 8R R z d zln d z R Q zln d d d dln d d 8R R R 6Qd 4R ln d R zd z d z ln d z R R d z d d zd d z zd
4 Prolem Consider sphere of rdius R. The potentil on the surfce of the sphere vries s (see figure elow) (free region inside nd outside) cos ) Compute the potentil inside nd outside of the sphere. ) Compute the electric field inside the sphere. c) Using Guss; lw, show tht while the electric field inside the sphere is non-zero, no chrge is contined inside the sphere. Solution:
5 ) Since the prolem hs n ziumuthl symmetry, we hve r, A l r l B l r l P l cos l Using tht we otin P cos P cos cos And thus cos P cos P cos P cos dxr,xpm x dx P cos P cosp m x m m, m m, 4 5 m, m, If we wnt to evlute the potentil inside of the sphere, we need to set B l nd otin nd thus for m nd for m nd thus dxr,xpm x dx l A m R m m 4 5 A R 5 A R A A A l R l P l x P m x r, P cos r R P cos
6 For the potentil outside of the sphere, we set A l nd otin nd thus for m dxr,xpm x dx l B m R m m B l R l P l cos P m x nd for m nd thus 4 5 B R 5 B R B R B R r, R r P cos R r P cos ) The electric field inside the sphere is then given y E r, r r, r, r r r 4 r R P cos r r sincos R 4 r R P cosr r R sincos c) Using Guss lw inside the shere E da 4 Thus, no chrges re contined inside the sphere. r R r d dcosp cos Prolem ) Consider the two conducting spheres with rdii nd ( ) s shown in the figure
7 elow. The volume etween the two spheres (region II) is filled with mteril of permitivity. The permitivity in regions I nd III is tht of free spce,. The two spheres re uniformly chrged with totl chrge Q. (i) Compute the mgnitude nd direction of the electric field in regions I, II, nd III. (ii) Compute the cpcitnce of the two spheres. ) Consider next two infinitely long concentric cylinders, s shown in the figure elow. The inner cylinder of rdius is conductor with liner chrge density. The second cylinder with inner rdius nd outer rdius c consists of mteril with permitivity nd is uniformly chrged with line chrge density ( ). The spce etween the two cylinders (i.e., r ) is filled with medium of permitivity. The medium outside the outer cylinder possesses the permitivity. Compute the potentil difference etween point t r c (mesured from the center of the inner cylinders) nd the center of the inner cylinder.
8 Solutions: ) (i) Using Guss lw, we cn compute the electric fields of the two sphere system. Region I: The electric field in reggion I for r,with r mesured from the center of the sphere, is given y E ds E4r q encl E the electric field is zero inside the inner sphere. Region II: The electric displcement field for r,with r mesured from the center of the sphere is given y D ds D4r Q D Q 4r r E D Q 4r r The electric field points rdilly inwrd. Region III: The electric field in reggion III for r,with r mesured from the center of the sphere, is given y E ds E4r q encl E the electric field is zero inside the outside the outer sphere.
9 (ii) The potentil difference etween the two sphere is nd the cpcitnce is thus Δ E tot dr Q 4 C Q Δ Q 4 4 r dr )We need to compute the electric field in the different regions of the prolem. i) r. WehveE inside the inner cylinder is zero, since it is conductor. ii) r. Weuse nd thus D d l rld l D E D r Note tht since the electric field points rdilly outwrds. iii) r c. Note tht the mteril in this region is insulting nd uniformly chrged. The volume chrge density is r c nd thus D d r c D r l rld r c r c l nd thus E D r r c iv) c r. Herewehve E r
10 We cn now compute the potentil difference Δ c c E tot dr c Edr c Edr Edr Edr where nd c c c Edr c r dr ln c Edr c c r r c r c c dr c ln c 4 r c dr nd nd Edr r dr ln Edr Thus I otin Δ ln c ln c ln Prolem 4 A solenoid of finite length nd redius hs N turns per unit length nd crries current I, with ciculr cross section s shown in the figure elow.
11 ) Compute the mgnetic induction on the solenoid xis in the limit N in terms of the ngles nd. ) For, how does the mgnetic induction scle with? Solutions: ) We first compute the mgnetic induction due to single loop. Using Biot-Svrt where db 4 I dl r r r z By symmetry, integrtion over the loop yields mgnetic induction long the z-xis. We then hve B 4 I dl r r 4 I z / dlr sinẑ where nd thus sin z /
12 B 4 I z / z / I z / ẑ ẑ z / Next, we consider the entire solenoid. B ẑ NI z dz z / z dz z / ẑ NI z z / ẑ NI cos cos The limit N is necessry to perform the integrtion over z. z z / ) From the ove expression, we hve limb limẑ NI z ẑ NI z z / z ẑ I N z z / Prolem 5 An electromgnetic plne wve is incident perpendiculr to lyered interfce, s shown in the figure elow. The indices of refrction of the three medi is n,n n nd n 4n while the permeility of ll three regions is the sme,. The thickness of the intermedite lyer is d. Ech of the other medi is semi-infinite.
13 d E B k n n n z= z=d ) Stte the oundry conditions t oth interfces in terms of the electric fields. ) Compute the rtio etween the incident electric field in medium nd the trnsmitted electric field in medium, i..e, compute E i /E t. c) If the thickness d is vried, the rtio E i /E t oscilltes. Wht is the period of the oscilltion? For which vlues of d is E i /E t the smllest? Solutions: ) Stte the oundry conditions t oth interfces in terms of the electric fields. The EM wve contins only components tht re perpendiculr to the interfce. In region, there is n incoming nd reflected wve, in region there is right-moving nd left-moving wve, nd in region, there is only trnsmitted wve. Thus the oundry conditions t z re E i E r E E E i E r c nd t z d we hve E E c E i E r n n E E E E E e ik d E e ik d E t e ik d E e ik d E e ik d c E te ik d c E e ik d E e ik d n n E t e ik d E t e ik d ) Compute the rtio etween the incident electric field in medium nd the trnsmitted electric field in medium, i..e, compute E i /E t. From the lst two equtions, I otin E n n E n n E t e ik d e ik d E te ik d e ik d E t e ik d e ik d E te ik d e ik d nd from the first two equtions
14 E i n n E n n E E E n n n n E t e ik d e ik d n n n n 9 E te ik d e ik d E te ik d e ik d E t e ik d e ik d nd thus 4 E i e E ikd e ik d n t n n n n n n n e ik d e ik d 9 e ik d e ik d nd hence 6 E i E t n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n 4 n n n n 4 n n 4 4 4sin k d 6sin k d e ik d n n n n sin k d e ik d cosk d n n sin k d or lterntively E i E t 4 n n n n n n n n sin k d sin k d 4 5 9sin k d c) If the thickness d is vried, the rtio E i /E t oscilltes. Wht is the period of the oscilltion? Assuming n n n, for which vlues of d is E i /E t the smllest? The period of the oscilltion is,nd E i /E t is the smllest for d m / with m eing n integer.
15 Mthemticl Formule Definitions r 4 d r r r r E r r B r 4 d r J r r r r r A r 4 d r J r r r Δ E r dr C Q Δ ; E ; B n In sphericl coordintes E B t ; E r,, r r,, r B J r r,, rsin r,, Integrls, Series, Expnsions nd Identities d cos K where K is the complete elliptic integrl
16 x x dx / / dx x x / x / dr r z r r z r z ln r z r c dx x x / x c c c dx Pl x l for even l for l ll!! l l for odd l dxpl x l dx Pl x dxpl xp m x l lm dx Pl x l P cos P cos cos P cos cos P cos 5cos cos r, A n r n B n r n P ncos n r dr r r dr lnr x x...
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