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Determine the electric field (magnitude and direction) at the point A (8.00 nm, 6.00 nm) caused by a particle located at the origin and carrying a charge of 7.00 μC .

Respuesta :

mshahzaib247

Answer:

E1 = 9.83 x [tex]10^{20}[/tex] [tex]NC^{-1}[/tex]

E2 = 1.748 x [tex]10^{21}[/tex] [tex]NC^{-1}[/tex]

Explanation:

E1 = k Q/r2 = 8.99 x [tex]10^{9}[/tex] x 7 x [tex]10^{-6}[/tex] / 8 x [tex]10^{-9}[/tex] x 8 x [tex]10^{-9}[/tex] = 9.83 x [tex]10^{20}[/tex] [tex]NC^{-1}[/tex]

E2 = k Q/r2 = 8.99 x [tex]10^{9}[/tex] x 7 x  [tex]10^{-6}[/tex] / 6 x  [tex]10^{-9}[/tex] x 6 x [tex]10^{-9}[/tex] = 1.748 x [tex]10^{21}[/tex] [tex]NC^{-1}[/tex]

The direction of the electric field will be from E1 to E2...

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