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?

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  • Anonymous
    1 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.

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  • 1 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

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