La asked in Science & MathematicsPhysics · 3 months ago

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In a vacuum, two particles have charges of q1 and q2, where q1 = +3.4C. They are separated by a distance of 0.36 m, and particle 1 experiences an attractive force of 5.0 N. What is the value of q2, with its sign?

1 Answer

  • 3 months ago
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    F = kQ₁Q₂/r²

    5 = (9e9)(3.4)Q / 0.36²

    Q = 5•0.36² / (9e9)(3.4) = 2.12e-11 C or 21.2 pC

    since the force is attractive, Q = –21.2 pC

    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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