Question

What is the force, in terms of Coulomb's constant, between two electrical charges of `22 C` and `-32 C` that are `6 m` apart?

Answer 1

It is an attraction force of `1.76*10^11 N`.

Explanation:

Coulomb's Law, in equation form, is

`F = k*(q_1*q_2)/r^2`

where F is the magnitude of the force of attraction, or repulsion, between 2 charges,
k is known as Coulomb's constant and has a value of `8.99×10^9 (N m^2)/C^2`,
`q_1 and q_2` are the 2 charges, in Coulombs,
and r is the distance between the charges, in meters.

The force is a force of attraction if F has a negative value and repulsion if the value of F has a positive value.

`color(green)(Edit " The green lines between here and the red text below replace those red lines.")`

`color(green)("Plugging the values into the formula")`
`color(green)(F = k * (22 C*(-32 C))/(6 m)^2)`
`color(green)(F = -k * 19.6 N)`

`color(green)("So it is an attraction force of " k * 19.6 N)`.

`color(red)("Plugging the values into the formula")`
`color(red)(F = (8.99×10^9 (N m^2)/C^2) * (22 C*(-32 C))/(6 m)^2)`
`color(red)(F = -175.8*10^9 N = -1.76*10^11 N)`

`color(red)("So it is an attraction force of " 1.76*10^11 N)`.

My apologies for the need to edit, I originally overlooked the request that the answer be in terms of k."

I hope this helps,
Steve