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A charge of -2.73 μC is fixed in place. From a horizontal distance of 0.0465 m, a particle of mass 9.50 x 10-3 kg and charge -9.56 μC is fired with an initial speed of 97.2 m/s directly toward the fixed charge. What is the distance of closest approach?

  • physic -

    KE=PE
    mv²/2=k•q1•q2/r

    r=2•k•q1•q2/ mv² =
    =2•9•10⁹•2.73•10⁻⁶•9.56•10⁻⁶/9.5•10⁻³•97.2²=
    =5.23•10⁻³ m

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