Question

The magnitude of the electrical force acting between a +2.4 x 10- C charge and a +1.8 x 10- C charge that are separated

Answer 1

Answer: 6.07 N

Explanation:

According to Coulomb's Law:  

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$  

Where:

$F_{E}$  is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant  

$q_{1}=2.4(10)^{-8} C$ and $q_{2}=1.8(10)^{-6} C$ are the electric charges

$d=0.008 m$ is the separation distance between the charges

Solving:

$F_{E}= 8.99(10)^{9} Nm^{2}/C^{2}\frac{(2.4(10)^{-8} C)(1.8(10)^{-6} C)}{(0.008 m)^{2}}$  

$F_{E}=6.07 N$