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aliina [53]
3 years ago
7

Um elétron é lançado entre duas placas eletrizadas como mostra a figura. Sejam v= 6x10^6 m/s, ângulo 45°, E= 2x10^3 N/C, d= 3 cm

, L=12cm, m=10^-30 kg, e=1,6x10-19 C. Despreze a ação gravitacional.
a) o eletron atingirá a placa negativa?
b) determinar a posicao em que o eletron atinge uma das placas.

Physics
1 answer:
Svetlanka [38]3 years ago
3 0
B hdbdudegjegedbfdrjf
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θ = 14.45º  = 0.252 rad

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The expression that describes the phenomenon of diffraction is

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  θ = sin-1 (0.24956)

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We pass radians

   θ = 14.45 (pi rad / 180º) = 0.252 rad

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