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?

Answer : Electric field, E = 50 N/C

Explanation :

It is given that,

Charge, q = +0.08 C

The force exerted by the electric field, F = 4 N

The electric field is defined as the ratio of electric force and electric charge.

Mathematically, it can be written as :

The magnitude of electric field is 50 N/C.

We know that the direction of electric force and the electric field are in same direction.

So, the direction of electric field is same as electric charge i.e. in right direction.                                                                                        

are there answer choices to this question? this seems like a question where you have options to choose from.

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