Lect02 - # $ )' * ' # + % % &' # ' * # ' ' # (...

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Unformatted text preview: # $ )' * ' # + % % &' # ' * # ' ' # ( " " ! ! " ! ! " ! " " ! , -. - * /0 1 F1 33 kq1q2 ^ r12 2 r12 kq1q3 ^ r13 2 r13 kq1q4 ^ r14 2 r14 E1 F1 q1 kq2 ^ r 2 12 r12 kq3 ^ r 2 13 r13 kq4 ^ r 2 14 r14 ' ' 2 # E F q ' E E i Q ^ k 2r r Qi ^ k 2 ri ri * E4 E2 E % 0% ' # * 2 ! E3 " 08 ! "I don't completely understand how to determine the direction of the electric field." field." 3 4) B? 09 # " 3 E E 12 4 % % % 2 % '6 * ' - 6 2 &7 ( &- 7 ( & 7( &7 ( E &- 7 ( & 7( 6 8 ' ' % * * ' # # ( ( ( ( 22 9 9 % 5 - - , , , 23 : 3 A B C D Force on q from +Q: Force on q from +2Q: Total force on q : F(from +Q) = + k Qq/r2 Qq/r (Acts to the right.) (Acts to the left.) So it acts to the right. F(from +2Q) = -k 2Qq/(2r)2 = + k Qq/2r2 Qq/2r + k Qq/r2 -k 2Qq/(2r)2 Qq/r 12 ; 6 # +q 8 ' ' = ' P d -q d Need to +q &0 ( &9 ( & ( E 0 & ( know d & ( Need to know d & q 2 E i Qi ^ k 2 ri ri q 2d 2 Ex k q d2 cos 4 Ey 20 k q d2 q 2d 2 sin 4 < ## # & ' % * ( E i Qi ^ k 2 ri ri ) * * ' # E 8 10 A & dq ^ k 2r r 1) ( % ' / 0 B == 6 # pt for E r charges 25 >?@ dE dq = dx 7 C C C D & >? @ ( # @# ' & ? @0 ( # @# # & > ? @ D( ! # @# # Asphere Vsphere E # Acylinder Vcylinder 4 R2 4 3 2 RL R2L R3 8 0( 0 9( ( 0 9 0( # %@ %@ 9( 4 3 =2 # # * # * # ' 2 4 3 > > @# ! @# QA V R3 QB 28 A 4 R2 QA QB R3 2 4 R 1 3 R 3 10) 29 y ' 6- 6 6# &6 ( > & 8 3 % ' # ' (= # 6> 2 @# 2 8 ' dq= dx dq= x r P h x a E 8 &0 ( dq ^ k 2r r dq r dx x 2 2 = dx &9 ( 2 a h2 & ( dx a 2 h 2 & ( dx (a x) 2 h 2 & ( dx x2 33 ! dE y ' 6- 6 6# &6 ( > & ' # ' (= 1 2 P 2 # 6> 2 @# 2 8 ' r h dE x x 8 3 % a dq= dx dq= x E 8 dq ^ k 2r r dE x = dq r2 dx (a x ) 2 h 2 Ex dE x &0 ( dE cos 1 &9 ( dE cos 2 & ( dE sin 1 & ( dE sin 2 33 " dE y ' 6- 6 6# &6 ( > & 8 ' # ' (= x 3 % # 6> 2 @# 2 8 ' P 2 dE x r h 1 2 a dq= dx dq= x E 8 dq ^ k 2r r Ex = 2 dq r2 dx (a x ) 2 h 2 Ex dE x dE cos 2 &0 ( k cos dx ( a x )2 h 2 &9 ( k cos dx 2 ( a x )2 h 2 0 a & ( none of the above 33 cos 2 DEPENDS ON x !! 4 dE y ' 6- 6 6# &6 ( > & ' # ' (= 1 2 P 2 # 6> 2 @# 2 8 ' r h dE x x 8 3 % a dq= dx dq= x E 8 &0 ( dq ^ k 2r r cos 2 = x a 2 dq r2 dx (a x ) 2 h 2 Ex dE x dE cos 2 h 2 &9 ( a x (a x ) 2 h 2 & ( a a 2 h 2 & ( a (a x ) 2 h 2 33 5 dE y ' 6- 6 6# &6 ( > & ' # ' (= 1 2 P 2 # 6> 2 @# 2 8 ' r h dE x 8 3 % E dx dq r2 8 dq ^ k 2r r Ex dE x x dq= dx dq= a x ( a x) 2 h 2 dE cos 2 cos 2 a x (a x) 2 h 2 E x (P) = a Ex ( P ) k 0 dx a x ( a x )2 h 2 3/ 2 Ex ( P ) k 1 h h h2 a 2 33 : dE y ' 6- 6 6# &6 ( > & ' # ' (= 1 2 P 2 # 6> 2 @# 2 8 ' r h dE x B % # # # ' 2 x dq= dx dq= a x Ex ( P ) k 1 h h h2 a 2 Ex ( P ) k 1 sin h 1 6 * F 33 ' % # 6 # ' Ex ( P ) k h /2 d cos 1 ; B C ' C 1 # % C , 3 ! 8 ; 77 0 / 8 G 4 < ...
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