Question

What is the force between 2 charged object that are 0.25 m apart? The charges have a value of
4.5 x 10-4 C and 5.9 x 10-5 ||

Answer 1

Answer:

3823 N

Explanation:

The magnitude of the electric force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=4.5\cdot 10^{-4}C$ is the charge on the 1st object

$q_2=5.9\cdot 10^{-5}C$is the charge on the 2nd object

r = 0.25 m is the distance between the objects

Therefore, the electric force between them is:

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-4})(5.9\cdot 10^{-5})}{(0.25)^2}=3823 N$