Question

A charge 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

Answer 1

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

Answer 2

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 :

$E=\dfrac{F}{q}$

$E=\dfrac{4\ N}{0.08\ C}$

$E=50\ N/C$

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

$E=50\ N/C$ in right direction.

So, the correct option is (D).