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

Question

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

Answer 1

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

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• NCS

  Lv 7

  3 months agoReport

  I meant to ask if you had remembered to convert.