Kelly asked in Science & MathematicsPhysics · 3 months ago

A point charge q= -0.45 nC is fixed at the origin.?

Part A-Where must a proton be placed in order for the electric force acting on it to be exactly opposite to its weight? (Let the y axis be vertical and the x axis be horizontal.)

For part A, I know the proton must be placed below the q.

Part B -Express your answer using two significant figures.

r =___km?

I need help with the formula values because none of my answers are making sense.

q1=|-0.45nC|=|0.45*10^-9C|

F=m*g

[K|q1||q2|]/r^2=m*g

1 Answer

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  • NCS
    Lv 7
    3 months ago
    Favorite Answer

    • For part A, I know the proton must be placed below the q.

    Right!

    • F=m*g

    • [K|q1||q2|]/r^2=m*g

    Yep. I'd write it as

    weight = electrostatic force

    m*g = kQq / d²

    and just plug in values:

    1.67e-27kg * 9.8m/s² = 8.99e9N·m²/C² * 1.6e-19C * 0.45e-9C / d²

    solves to

    d = 6300 m = 6.3 km

    If you find this helpful, please award Best Answer. You get points too!

    • NCS
      Lv 7
      3 months agoReport

      I meant to ask if you had remembered to convert.

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