Question

A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an electric field. What are the magnitude and direction of the electric field at the point in question

Answer 1

Answer:

$E = -6 \ N/C$

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

Explanation:

From the question we are told that

   The charge on the small object is $Q = -4.00 \ nC = -4.00 *10^{-9} \ C$

   The force is  $F = 24 \ nN = 24 *10^{-9} \ N$

    Generally the magnitude of the electric  field is mathematically represented as

       $E = \frac{F}{Q}$

=>    $E = \frac{ 24 *10^{-9}} {-4 *10^{-9 }}$

=>     $E = -6 \ N/C$

Generally given that the electric field is negative it mean that its direction is opposite to that of the force