Question

4) Duas cargas elétricas com valores respectivos de 4.10-6 C e 8.10-6 C interagem com força de 40 N. Determine a distância entre estas cargas elétricas, sendo K = 9.109(SI)

Answer 1

Dados:

$q_1=4\times10^{-6}\ C\\ q_2=8\times10^{-6}\ C\\ F=40\ N\\ k_0\approx9\times10^9\ Nm^2/C^2$

Explicação:

$\boxed{F=k_0\dfrac{q_1q_2}{d^2}}\ \therefore\ d^2=k_0\dfrac{q_1q_2}{F}\ \therefore\ \boxed{d=\sqrt{k_0\dfrac{q_1q_2}{F}}}$

Substituindo os valores, teremos:

$d=\sqrt{9\times10^9\bigg(\dfrac{4\times10^{-6}\big(8\times10^{-6}\big)}{40}\bigg)}=$

$=\sqrt{\dfrac{36\times10^{-3}}{5}}=6\sqrt{2\times10^{-4}}\ \therefore\ \boxed{d=6\sqrt{2}\times10^{-2}\ m}$

Answer 2

Resposta:

$f = k\frac{q1.q2}{ {d}^{2} }$

$40= 9 \times {10}^{9} \frac{4 \times {10}^{ - 6} .8 \times {10}^{ - 6} }{ {d}^{2} } \\ 40 {d}^{2} = 9 \times {10}^{9} \times 32\times {10}^{ - 12} \\ {d}^{2} = \frac{288 \times {10}^{ - 3} }{40} \\ {d }^{2} = 72 \times {10}^{ - 4} \\ d = \sqrt{72 \times {10}^{ - 4} } \\ d = 8.48 \times {10}^{ - 2} \\ d = 8.48cm$