Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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.

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.