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trasher [3.6K]
3 years ago
5

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

Physics
1 answer:
Brrunno [24]3 years ago
3 0

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  

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Divide each side by (0.9):

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Divide each side by (9.8 m/s²):

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