Question

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Answer 1

4_
Pela 1ª Lei de Coulomb, temos:
$F = K \frac{Qq}{d^{2} }$
$0,1 = 9.10^{9} \frac{(4.10^{-8}).4.10^{-8} }{d^{2} }$
$0,1 = 9.10^{9} \frac{16.10^{-16} }{d^{2} }$
$0,1 = \frac{144.10^{-7} }{d^{2} }$
$d^{2} = 144.10^{-6}$
$d = \sqrt{144.10^{-6} } = 12.10^{-3} = 0,012$
Portanto, letra C)

5_
$F = K \frac{Qq}{d^{2} }$
$16.10^{-1} = 9.10^{9} \frac{(2.10^{-6})q}{(0,3)^{2} } \\ 16.10^{-1} = 9.10^{9} \frac{(2.10^{-6})q}{9.10^{-2} } \\ 16.10^{-1} = 2.10^{5}q \\ q = \frac{16.10^{-1} }{2.10^{5} } = 8.10^{-6}C$

6_
$F = K \frac{Qq}{d^{2} } \\ F = 9.10^{9} \frac{(5.10^{-6}).5.10^{-6} }{5.10^{-1} } \\ F = 9.10^{9}.(5.10^{-11}) \\ F = 45.10^{-2}C$