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

r = 5.335 meters

Step-by-step explanation:

Given that,

Charge 1,
`q_1=-165\ \mu C`

Charge 2,
`q_2=115\ \mu C`

Force of attraction between two charges, F = 6 N

The force of attraction between two charges is given by :

`F=k(q_1q_2)/(r^2)`, r is the separation between two charges

`r=\sqrt{(kq_1q_2)/(F)}`

`r=\sqrt{(9* 10^9* 165* 10^(-6)* 115* 10^(-6))/(6)}`

r = 5.335 m

So, the separation between two charges is 5.335 meters. Hence, this is the required solution.