contestada

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 _________.

Respuesta :

Answer:

F = 311*10^{-5} N

Explanation:

given data:

[tex]q_1 = 94 nc[/tex]

[tex]q_2 = 53 nc[/tex]

distance between the charges r =12 cm

force  between the charges is given by

[tex]F = \frac{q_2*q_2}{u\pi \epsilon_o r^2}[/tex]

[tex]\frac{1}{u \pi \epsilon_o}[/tex] = 9*10^9 Nm2/c2

[tex]F = \frac{94*10^{-9}*(53*10^{-9})*9*10^9}{(12*10^{-2})^2}[/tex]

F = 311*10^{-5} N