Question

1. What is the magnitude of the force on a charge of +40 μC that is 0.6 m from a charge of - 80 μC?

Answer 1

The magnitude of the force is 79.893 N.

Step-by-step explanation:

The magnitude of the electrostatic force between the two particles is determined by Coloumb's Law, whose formula is:

`F = (\kappa \cdot |q_(A)|\cdot |q_(B)|)/(r^(2))` (1)

Where:

`\kappa` - Electrostatic constant, in newtons-square meters per square coulomb.

`q_(A), q_(B)` - Electric charges, in coulombs.

`r` - Distance, in meters.

If we know that
`\kappa = 8.988* 10^(9)\,(N\cdot m^(2))/(C^(2))`,
`q_(A) = +40* 10^(-6)\, C`,
`q_(B) = - 80* 10^(-6)\, C` and
`r = 0.6\,m`, then the magnitude of the force is:

`F = (\kappa \cdot |q_(A)|\cdot |q_(B)|)/(r^(2))`

`F = 79.893\,N`

The magnitude of the force is 79.893 N.