Question

What is the force, in terms of Coulomb's constant, between two electrical charges of `24 C` and `9 C` that are `2 m` apart?

Answer 1

`F=4.9086×10^-7N`

Explanation:

Using Coulomb's law
`F=(KQ_1Q_2)/d^2`
Given
`Q_1=24c`
`Q_2=9c`
`d=2m`
Constant,`K=9.09×10^-9Nm^2c^-2`
Putting in values in the above eqn
`F=[(9.09×10^-9(24×9))/2^2]N`
`=>[(9.09×10^-9(216))/4]N`
`=>[(1.96344e-6)/4]N`
`F=4.9086e-7N~~5×10^-7N`

Answer 2

`k*54 \ "C/m"^2`

Explanation:

Coulomb's law states that,

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

• `q_1,q_2` are the charges of the two objects in coulombs

• `r` is the distance between them in meters

And so, we got:

`F=k*(24 \ "C"*9 \ "C")/(2 \ "m")^2`

`=k*(216 \ "C"^2)/(4 \ "m"^2)`

`=k*54 \ "C/m"^2`