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

`E = -6 \ N/C`

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

Step-by-step 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 = (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