Question

Two particles of electrical charges Q1=3.8×10-⁶C and q,=4.4×10-⁶C are separated in vacuum by a distance of 4.0.10-⁸ m. Since K=9.0.10⁹ N.m²/C², the intensity of the interaction force between them, in newtons, is?

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Answer to a math question Two particles of electrical charges Q1=3.8×10-⁶C and q,=4.4×10-⁶C are separated in vacuum by a distance of 4.0.10-⁸ m. Since K=9.0.10⁹ N.m²/C², the intensity of the interaction force between them, in newtons, is?

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Frederik
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101 Answers
Na lei de Coulomb, a fórmula para a intensidade da força de interação entre duas cargas elétricas é dada por:

F = \frac{K \cdot |q_1 \cdot q_2|}{r^2}

Onde:
- F é a intensidade da força de interação,
- K é a constante eletrostática,
- q1 e q2 são as cargas elétricas das partículas,
- r é a distância entre as partículas.

Substituindo os valores fornecidos na fórmula, temos:

F = \frac{(9,0 \times 10^9) \cdot |(3,8 \times 10^{-6}) \cdot (4,4 \times 10^{-6})|}{(4,0 \times 10^{-8})^2}

Simplificando os valores:

F = \frac{(9,0 \times 3,8 \times 4,4) \times (10^9 \times 10^{-6} \times 10^{-6})}{(4,0)^2 \times (10^{-8})^2}

F = \frac{119,52 \times 10^{-1} \times 10^{-1}}{16 \times 10^{-16}}

F = \frac{119,52}{16} \times \frac{10^{-1} \times 10^{-1}}{10^{-16}}

F=9.405\times10^{13}

Portanto, a intensidade da força de interação entre as duas partículas é de aproximadamente 9.405 × 10^13 N.

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