Eletromagnetismo II Aula 20 ExercÃcios: faça os problemas ... - IFSC
Eletromagnetismo II Aula 20 ExercÃcios: faça os problemas ... - IFSC
Eletromagnetismo II Aula 20 ExercÃcios: faça os problemas ... - IFSC
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concluím<strong>os</strong> que há, no máximo, um instante de tempo em que a Eq. (1) vale edefinim<strong>os</strong> esse instante como o tempo retardado:t R = t − |r − r 0 (t R )|.cO resultado da integral acima pode ser escrito em term<strong>os</strong> do tempo retardadocomoˆ +∞−∞dt ′ δ (f (t ′ ))|r − r 0 (t ′ )|==ˆ +∞−∞dt ′ δ (t ′ − t R )|r − r 0 (t ′ )| ∣1∣|r − r 0 (t R )|∣ df(t′ )∣ df(t′ )dt ′dt ′ ∣ ∣∣t′=t R.∣ ∣∣t′=t RCalculem<strong>os</strong>, explicitamente, a derivada temporal da função f:df (t ′ )= d (t ′ − t + |r − r 0 (t ′ ))|dt ′ dt ′ c= 1 + 1 d |r − r 0 (t ′ )|.c dt ′Notem<strong>os</strong> quePortanto,Comod |r − r 0 (t ′ )|dt ′ =ˆ +∞−∞=dt ′ δ (f (t ′ ))|r − r 0 (t ′ )|1 d |r − r 0 (t ′ )| 22 |r − r 0 (t ′ )| dt ′12 |r − r 0 (t ′ )|d [r − r 0 (t ′ )] · [r − r 0 (t ′ )]dt ′= [r − r 0 (t ′ )]|r − r 0 (t ′ )| · d [r − r 0 (t ′ )]dt ′= − [r − r 0 (t ′ )]|r − r 0 (t ′ )| · dr 0 (t ′ )dt ′= − [r − r 0 (t ′ )]|r − r 0 (t ′ )| · v (t′ ) .=1∣∣|r − r 0 (t R )| − [r − r 0 (t R )] · v(t R).∣∣ v ∣ < 1,cc4
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