Question

Which calculates the intensity of an electric field at a point where a 0.50 C charge experiences a force of 20. N?

Answer 1

Answer:
`40\ N/C`

Step-by-step explanation:

Given

Magnitude of charge is
`q=0.5\ C`

Force experienced is
`F=20\ N`

Electric field intensity is the electrostatic force per unit charge

`\therefore E=(F)/(q)\\\\\Rightarrow E=(20)/(0.50)\\\\\Rightarrow E=40\ N/C`

Thus, the electric field intensity is
`40\ N/C`