Question

The magnitude of the electrical force acting between a +2.4 x 10- C charge and a +1.8 x 10- C charge that are separated

by 0.008 m is
N, rounded to the tenths place

Answer 1

Answer: 6.07 N

Step-by-step explanation:

According to Coulomb's Law:

`F_(E)= K(q_(1).q_(2))/(d^(2))`

Where:

`F_(E)` is the electrostatic force

`K=8.99(10)^(9) Nm^(2)/C^(2)` is the Coulomb's constant

`q_(1)=2.4(10)^(-8) C` and
`q_(2)=1.8(10)^(-6) C` are the electric charges

`d=0.008 m` is the separation distance between the charges

Solving:

`F_(E)= 8.99(10)^(9) Nm^(2)/C^(2)((2.4(10)^(-8) C)(1.8(10)^(-6) C))/((0.008 m)^(2))`

`F_(E)=6.07 N`