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Three charges sit on the vertices of an equilateral triangle, the side of which are 30.0 cm long. If the charges are A = +4.0 C, B = +5.0 C, and C = +6.0 C (clockwise from the top vertex), find the force on each charge.

  • physics - ,

    Charge A(q1)
    F12 = k•q1•q2/a².
    F13 = k•q1•q3/a².
    F(A) = sqrt(F12² + F13² - 2•F12•F13•cos 120º).

    Charge B(q2)
    F21 = k•q1•q2/a².
    F23 = k•q2•q3/a².
    F(B) = sqrt(F21²+F23²-2•F21•F23•cos120º).

    Charge C (g3),
    F31 = k•q1•q3/a².
    F32 = k•q2•q3/a².
    F(C) = sqrt(F31² + F32² - 2•F31•F32•cos 120º).

  • physics - ,

    How did you get F12 and F13 ??

  • physics - ,

    physics - sara, Friday, August 17, 2012 at 10:45pm
    How did you get F12 and F13 ??

    F12 = k•q1•q2/a²=9•10^9•4•10^-6•5•10^-6/0.09=2 N
    F13 = k•q1•q3/a² = 9•10^9•4•10^-6•6•10^-6/0.09=2.4 N

    F(A) = sqrt(F12² + F13² - 2•F12•F13•cos 120º)=
    =sqrt[4+5.76-2•2•2.4(-0.5)] =3.82 N

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