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.
I need help with the formula values because none of my answers are making sense.
- NCSLv 73 months agoFavorite Answer
• For part A, I know the proton must be placed below the q.
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
d = 6300 m = 6.3 km
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