Charles K. Alexander e Matthew Sadiku - Fundamentos De Circuitos Elétricos - 5a Edição

Problema 1.1 Quantos coulombs são representados pelas seguintes quantidades de elétrons?

(a) (6,482 \times 10^{17})
(b) (1,24 \times 10^{18})
(c) (2,46 \times 10^{19})
(d) (1,628 \times 10^{20})

Dado fundamental:
A carga de um elétron é:

e = -1,602 \times 10^{-19} \ \text{C}

A carga total Q de n elétrons é:

Q = n \times e

(a) n = 6,482 \times 10^{17}
Q = (6,482 \times 10^{17}) \times (-1,602 \times 10^{-19})
Q = -1,0384 \times 10^{-1} \ \text{C}

Q = -0,10384 \ \text{C} = -103,84 \ \text{mC}

Resposta (a): -103,84 \ \text{mC}

(b) n = 1,24 \times 10^{18}
Q = (1,24 \times 10^{18}) \times (-1,602 \times 10^{-19})
Q = -1,98648 \times 10^{-1} \ \text{C}

Q = -0,198648 \ \text{C} = -198,65 \ \text{mC}

Resposta (b): -198,65 \ \text{mC}

(c) n = 2,46 \times 10^{19}
Q = (2,46 \times 10^{19}) \times (-1,602 \times 10^{-19})

Q = -3,94092 \ \text{C}

Resposta (c): -3,941 \ \text{C}

(d) n = 1,628 \times 10^{20}
Q = (1,628 \times 10^{20}) \times (-1,602 \times 10^{-19})
Q = -26,08056 \times 10^{1} \ \text{C}?
Vamos calcular com cuidado:
1,628 \times 1,602 = 2,608056
10^{20} \times 10^{-19} = 10^{1}

Q = -2,608056 \times 10^{1} = -26,08056 \ \text{C}

Resposta (d): -26,08 \ \text{C}

Respostas:
(a) -103,84 \ \text{mC}
(b) -198,65 \ \text{mC}
(c) -3,941 \ \text{C}
(d) -26,08 \ \text{C}