Question

What would the electrostatic force be for two 0.005C charges 50m apart?

Answer 1

90 N

Step-by-step explanation:

The electrostatic force between two charges is given by:

`F=k(q_1 q_2)/(r^2)`

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem we have

q1 = q2 = 0.005 C

r = 50 m

So the electrostatic force is

`F=(9\cdot 10^9 N m^2 C^(-2))((0.005 C)^2)/((50 m)^2)=90 N`