contestada

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
by 0.008 m is
N, rounded to the tenths place

Respuesta :

cryssatemp

Answer: 6.07 N

Explanation:

According to Coulomb's Law:  

[tex]F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}[/tex]  

Where:

[tex]F_{E}[/tex]  is the electrostatic force

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

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

[tex]d=0.008 m[/tex] is the separation distance between the charges

Solving:

[tex]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}}[/tex]  

[tex]F_{E}=6.07 N[/tex]  

Otras preguntas