Vacuum/attractive forces physics problem?
In a vacuum, two particles have charges of q1 and q2, where q1 = +4.6C. They are separated by a distance of 0.38 m, and particle 1 experiences an attractive force of 3.1 N. What is the value of q2, with its sign?
- Anonymous1 month ago
I'm guessing the given charge is actually 4.6 µC. That would give q2 a charge of -1.08e-5 C ≈ -11 µC.
- billrussell42Lv 71 month ago
3.1 = (9e9)(4.6)Q₂/0.38²
Q₂ = 3.1•0.38² / (9e9)(4.6) = 1.08e-11 C
negative as they are attracted.
Coulomb's law, force of attraction/repulsion
F = kQ₁Q₂/r²
Q₁ and Q₂ are the charges in coulombs
F is force in newtons
r is separation in meters
k = 8.99e9 Nm²/C²