Respostas

2014-05-02T13:35:41-03:00
3 μC →  3.10^-6
16μC  →  16.10^-6

d →  2 m

k → 9.10^9


\boxed{F=K_{0}( \frac{|Q_1|.|Q_2|}{d^2} )}

F=9.10^9( \frac{3.10^{-6}*16.10^{-6}}{2^2} )

F= \frac{9.10^9*4,8.10^{-11}}{4} \\
\\F= \frac{43,2.10^{-2}}{4}

\boxed{F=1,08.10^{-1}~N}