Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

Physics: Coulombs Law: Two small, positively charged spheres have a combined charge of 13.0 × 10-5 C+++?

If each sphere is repelled from the other by an electrostatic force of 0.600 N when the spheres are 1.80 m apart, what is the charge on the sphere with the smaller charge?

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  • ?
    Lv 7
    1 month ago
    Favorite Answer

    0.600 N = (9 x 10^9 Nm^2/C^2)*Q*(1.3 x 10^(-5)C - Q) / (1.80 m)^2.

    Agreeing that "Q" will come out in coulombs, let's abandon the units for a moment.

    0.600 = (9 x 10^9) Q*(1.3 x 10^(-5)C - Q) / 1.80^2 =>

    1.944 = (9 x 10^9) Q*(1.3 x 10^(-5) C - Q) =>

    (9 x 10^9)Q^2 - 1.17 x 10^5 Q + 1.944 = 0.

    Now you have a quadratic equation to be solved for Q.

    Use the quadratic formula, that's what it's for !

  • 1 month ago

    13.0 × 10-5 = 130–5 = 125 C

    strange number,,,

    what does +++ indicate?

    F = kQ₁Q₂/r² = 0.8

    Q₁Q₂ = 0.8•1.8²/9e9 = 2.88e-10

    Q₁+Q₂ = 125

    Q₁ = 125 – Q₂

    Q₁Q₂ = 2.88e-10

    (125 – Q₂)Q₂ = 2.88e-10

    Q₂²  – 125Q₂ + 2.88e-10 = 0

    quadratic equation:to solve ax² + bx + c = 0x = [–b ± √(b²–4ac)] / 2aQ₂= [125 ± √(125²–4•2.88e-10)] / 2

    Q₂= [125 ± 124.999999999983] / 2

    Q₂ = 125 C

    Q₁ = 8.6e-12 C

    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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