Acharge of +0.08 c moves to the right due to a 4 n force exerted by an electric field. what is the magnitude and direction of the electric field? a) 0.32 n/c left b) 0.32 n/c right c) 50 n/c left d) 50 n/c right

This is late but the answer is 50 n/c right

Answer : Electric field is given by, E = 50 N/C in right.

Explanation :

It is given that,

Charge, q = 0.08 C (moves to the right)

Force exerted by the electric field, F = 4 N

We know that the electric field is defined as :

The direction of field is same as the direction of electric force i.e.

in right direction.

So, the correct option is (D).

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explanation:

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