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a (9x10^-5)C and (6x10^-4)C charge are 3 meters apart, how much force is created between the two?

Sagot :

According to Coulomb's Law, if two point charges q₁ and q₂ are placed a distance r apart, the magnitude of the force between them is given by:

[tex]F=k\frac{q_1q_2}{r^2}[/tex]

Where k is the Coulomb's constant:

[tex]k=8.99\times10^9N\frac{m^2}{C^2}[/tex]

Replace q₁=(9x10^-5)C and q₂=(6x10^-4)C, as well as r=3m to find the force between the two charges:

[tex]F=(8.99\times10^9N\frac{m^2}{C^2})\times\frac{(9\times10^{-5}C)(6\times10^{-4}C)}{(3m)^2}=53.94N\approx54N[/tex]

Therefore, the force between the two charges is approximately 54N.

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