Question

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

Answer 1

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:

`F=k(q_1q_2)/(r^2)`

Where k is the Coulomb's constant:

`k=8.99*10^9N(m^2)/(C^2)`

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

`F=(8.99*10^9N(m^2)/(C^2))*((9*10^(-5)C)(6*10^(-4)C))/((3m)^2)=53.94N\approx54N`

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