Electrostatic Energy of System - Electricity and Magnetism - Solved Exam, Exams of Electromagnetism and Electromagnetic Fields Theory

Download Electrostatic Energy of System - Electricity and Magnetism - Solved Exam and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! Solutions to PC 2131 AY0708 Sem 2 1. The electrostatic energy of the system is the work needed to bring the charges into this particular configuration. We have 1 2 3TE W W W   , and 1 0W  as there is no field originally. W2 and W3 respectively: 2 0 2 2 0 21 2 16)2(4 a q a qq W   2 0 2 2 0 2 2 0 23 2 0 13 3 24 2 44 a q a q a qq a qq W     Adding all the work together, 2 0 2 2 0 2 2 0 2 321 16 7 216 0 a q a q a q WWWET   2. Using Gauss’s Law for electric displacement, r r q D qrD adDdVdvD D i i ifi fi ˆ 4 )4( 2 2             and Ei and Pi respectively are: r r qD E ii ˆ 4 2   r r q EEEP iiiei ˆ4 )1()()1( 2 0 00 0 0        For outside the sphere, r r q DD io ˆ4 2  Since there is no dielectric material, 0oP r r qD EPED ooooo ˆ4 200 0    3(a) AB   Magnetic flux,   ldAadAadB  )( 3(b) AB   JAB  0)(  AAA  2)()(  To keep the magnetic potential the same, 0)(  A  JAAAAB  0 22)()(  Since this is a Poisson equation, the solution is dV r J A     4 0 National University of Singapore Physics Society 2009 4(a)(i) For r a , using Gauss’ law,   dVdaE 1 0 1   r r ElrrlE ˆ 2 )()2( 0 12 0 1       At ar  , r a E ˆ 2 0 1     4(a)(ii) For r > c, by Gauss’ law,   dVdaE )( 1 21 0   rbca r E bcllarlE ˆ))(( 2 1 )()()2( 22 2 2 1 0 22 0 22 0 1             4(b) E is 0 for cr  , 2 22 2 1 22 2 2 1 22 2 2 1 0 0)( 0ˆ))(( 2 1 a cb bca rbca r E           5 Energy of a capacitor, 2 1 2 W CV dx dC VCV dx d dx dW F 22 2 1 2 1       C is the sum of both the dielectric part and non- dielectric part, )1( 22 0 2 00 2        rr s wV w s xw s wx dx dV dx dW F    6 From Bio-Savart law,     2 0 2 0 2 0 cos 4 sin 4 ˆ 4 r dlI r dlI r rldI B         cos s r  tansl   dsdl sec2 Substituting into the equation, )sin(sin 4 cos 4 sec sec cos 4 cos 4 12 002 22 0 2 0 2 1                s I d s I s dsI r dlI B National University of Singapore Physics Society 2009                                                   2 3 2 2 2 2 3 220 cos2 cos cos2 )cos( 4 1      rD a Dr a D a D rQ rDrD DrQ            22 2 3 22 2 2 3 22 2 3 22 2 2 3 22 2 3 2 2 2 2 3 220 0 0 cos2 4 1 coscos cos2 4 1 cos2 cos cos2 )cos( 4 1 cos2 cos cos2 )cos( 4 1 Da rDaD Q a D a D Da rDaD Q aDDa D a D Q rDaD DaQ aD a Da a D a D aQ rDaD DaQ r V ar ind                                                                                                                              To find total induced charge,       ')2( 2))((2 ))(( 11 2 )( )( 1 )( 1 2 )( 2 1 2 1 2 )( 1 cos2 2 )( sincos2)2( 4 )( sincos2 4 )( 22 22 22 2222 22 0 2 1 22 22 2 3 222 22 22 3 22 22 Qa D Q aDaD aDaD D DaDaQ aDaDD DaQ aDaDD DaQ aDaDaDaDD DaQ aD rDaD DaQa drDaDa a DaQ ddarDaD a DaQ daQ indtotal                                                                            National University of Singapore Physics Society 2009 1(c)(i) The conductor surface has an induced charge of Qa D  on it. A second image charge is needed to make the net charge on the surface 0. Using the second configuration, adding ''Q on the center so the sphere will remain and equipotential, Net charge on conductor = 0 Q D a QQQQ  '"0"' 1(c)(ii) "Qoldnew VVV                                            Dr a D a rr D aD a rDrD Q r Q rbrb Q rDrD Q    cos2 1 cos2 1 4 4 " cos24 ' cos24 2 2 22220 0 22 0 22 0 D Q D aDDa D aD a aDaD Q Da a D a aa D aD a aDaD Q V arnew 0 22 22 0 2 2 22220 4 1 cos2 1 cos2 1 4 cos2 1 cos2 1 4                                                                       National University of Singapore Physics Society 2009 2(a)(i) For a conducting spherical shell, a Q V 04  a V Q C 04 2(a)(ii) Field energy density, 202 1 E dV dW  4 0 2 22 2 0 0 2 0 3242 1 2 1 r Q r Q E dV dW          2(a)(iii)Total field energy,  dVEEtotal 202 1    R Q r Q ddr r Q dddrr r Q E R Rtotal 0 2 0 0 2 02 0 2 2 2 42 0 2 2 0 8 cos 1 16 sin 1 2 32 sin 1 162                                       2(a)(iv)Work done, R Q EW total 0 2 8  2(b)(i) 2 04 r Q E   aR aRVQ Ra Q dr r Q ldEV a R         0 0 2 0 4 11 44    Electric field near the surface of the inner sphere, )()( )( 2 aRa RV raR aRV aE ar      2(b)(ii) )2( )( 22 aR aaR RV da dE    2 0 R a da dE  E is minimum when 2 R a  3(a)(i) )(000 MHBMHB    3(a)(ii) In a linear material, HM m   HHHHMHB mm    )1()()( 000 where )1(0 m  National University of Singapore Physics Society 2009