A charge of 0.05 C moves a negative charge upward due to a 2 N force exerted by an electric field. What is the
magnitude and direction of the electric field?
0.03 N upward
0.03 N downward
40 N upward
40 N downward

40 N/C, downward

Explanation:

- Magnitude:

The relationship between the force experienced by a charge and the electric field strength is given by

F=qE

where

F is the force

q is the charge

E is the electric field strength

In this problem, we have

q = 0.05 C is the charge

F = 2 N is the force exerted on the charge

By solving the formula for E, we find the magnitude of the electric field:

E=\frac{F}{q}=\frac{2 N}{0.05 C}=40 N/C

-Direction:

The direction of an electric field corresponds to the direction of the force that would be exerted on a positive charge immersed in the field. Therefore, for a negative charge, the electric field and the electric force have opposite directions.

Since the charge in the problem is negative, and since the charge is pushed upward (so, the force is upward), the electric field must be downward.

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thinking it would be d not too sure but i think it's right