Question

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in meter rounded to three decimal places? (k = 1/ 40 = 9.00 x 10°N.m2/C2, 1°C = 10 C)

Answer 1

• The distance between the charges is 5,335.026 m

Step-by-step explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

`F = k (q_1 q_2)/(d^2)`

where k is Coulomb's constant,
`q_1` and
`q_2` are the charges and d is the distance between the charges.

Working a little the equation, we can take:

`d^2 = k (q_1 q_2)/(F)`

`d = \sqrt{ k (q_1 q_2)/(F)}`

And this equation will give us the distance between the charges. Taking the values of the problem

`k= 9.00 \ 10^9 (N \ m^2)/(C^2) \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00`

(the force has a minus sign, as its attractive)

`d = \sqrt{ 9.00 \ 10^9 (N \ m^2)/(C^2) ((165.0 \mu C) (115.0 C))/(- 6.00 \ N)}`

`d = \sqrt{ 9.00 \ 10^9 (N \ m^2)/(C^2) ((165.0 \mu C) (115.0 C))/(- 6.00 \ N)}`

`d = √( 28,462,500 \ m^2)}`

`d = 5,335.026 m`

And this is the distance between the charges.