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A small object carrying a charge of is acted upon by a downward force of 20.0 nN when placed at a certain point in an electric field. (a) What are the magnitude and direction of the electric field at the point in question? (b) What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?

Respuesta :

Manetho

Answer:

a)  4 N/C

b) 6.4×10^{-19} N

Explanation:

Since, the magnitude of charge is not given i am assuming it to be as 5 nC positive.

therefore, magnitude of electric field: E = F/q = 20/5 = 4 N/C

direction: downward

b)  F = q E = 1.6 e^{-19}×4 = 6.4×10^{-19} N

upward

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