please help me with this hard electrostatics problem?

A point charge q = +0.71 nC is fixed at the origin. Where must a proton be placed in order for the electric force acting on the proton to be exactly opposite to its weight? (Let the y axis be vertical and the x axis be horizontal.)

I understand that it will be above q since it's a proton...but that's about all I know

Update:

T'hat's the answer I kept on getting, but that's not the right answer,

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  • 8 years ago
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    Force acting up on the proton is k q*e/r²---------------e is the charge of the proton

    This must equal the weight of the proton

    k q e/y²= mg

    y²= k q e/ mg

    y²= 9e9*0.71e-9*1.602e-19/ (9.109e-31*9.8)

    y = 3.39 e+5m

    ========================

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