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?

Question

asked Apr 26, 2017 156k views

2 Answers

FredTheLover · Answer 1 · 2017-05-03T04:16:25+0000

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 :

E=(F)/(q)

$E=(4\ N)/(0.08\ C)$

$E=50\ N/C$

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.