Question

Q4-Q5, urgente preciso muito

Answer 1

Resposta:

4- a) F= 9x10^9 . 2x10^-8 . 6x10^-8 / (20x10^-2)²

F = 108x10^-7 / 400x10^-4

F = 0,27 x 10^-3

F = 2,7 x 10^-4 N

b) Q1 = Q2

Q1+Q2/2

2.10^-8 - 6.10^-8/2

-2.10^-8

F= 9.10^9 . 2.10^-8 . 2.10^-8/ 1.10^-2

F= 36.10^-7/1.10^-2

F= 3,6.10 ⁻⁴ Newton

5- Feletrica= 6 N  

d= 3 m

Fel= Ko x Q1 x Q2/ d^2

6= 9.10^9 x 2Q x 3Q/3^2

Q= aproximadamente 3,16.10^-5, logo Q1= 2x 3,16.10^-5 e Q2= 3x 3,16.10^-5