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A positively charged particle initially at rest on the ground accelerates upward to 140 m/s in 2.10 s. The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform.What are the magnitude and direction of the electric field?

Respuesta :

Answer:

764.7 N/C, the direction is upward

Explanation:

E electric field of  the electron = F / charge

F force of the electron = Eq

acceleration of the body = change in velocity / time = 140 m / 2.10 s = 66.67 m/s² since the electron is from rest.

F upward = F of electron - force of gravity

F upward = Eq - mg

ma = Eq -mg

divide through with m

a = E (q/m) - g

(a + g) / (q/m) = E

(66.67 + 9.8) / 0.100 = 764.7 N/C, the direction is upward

ArcticPeeper

g and the electric field in this region is constant and uniform.What are the magnitude and direction of the electric field?



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