Question

A positive charge of 94.0 nC is 12.0 cm from a negative charge of 53.0 nC. The force on one of the charges due to the other is approximately _________.

Answer 1

F = 311*10^{-5} N

Step-by-step explanation:

given data:

`q_1 = 94 nc`

`q_2 = 53 nc`

distance between the charges r =12 cm

force between the charges is given by

`F = (q_2*q_2)/(u\pi \epsilon_o r^2)`

`(1)/(u \pi \epsilon_o)` = 9*10^9 Nm2/c2

`F = (94*10^(-9)*(53*10^(-9))*9*10^9)/((12*10^(-2))^2)`

F = 311*10^{-5} N