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Amazingboliver4518
04.03.2021 •
Physics
A point charge q = -0.41 nC is fixed at the origin. Where must an electron be placed in order for the
electric force acting on it to be exactly opposite to its weight? (Let the y axis be vertical with the positive
direction pointing up.)
e= 1.60•10^-19 C
me= 9.11•10^-31 kg
K= 8.99•10^9 N•m^2/C^2
Solved
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Ответ:
d = 286809.88 m
Distance can be found on one of the following coordinates which is given below
(d,0), (0,d), (-d,0), (0,-d)
Explanation:
As the electron and charge (q) are negatively charged they will repulsive
Let F(Force)= W (Weight of electron)
where:
Force (F) between the charges = kq₁q₂/d²
Weight of electron = mg
So:
F=W
kq₁q₂/d² = mgeq(i)
Where
mass of electron(m) = 9.11x10⁻³¹ kg
k = 9x10⁹ Nm²/C²
q₂ (charge of electron) = 1.6x10⁻¹⁹C
d =?
put values in eq(i) we get
{(9x10⁹Nm²/C²)(0.51nC)(1.6x10⁻¹⁹C)}/d^2 = (9.11x10⁻³¹ kg )(9.8 m/s²)
d² = 8.226x10¹⁰ m²
d =√(8.226x10¹⁰ m²)
d = 286809.88 m
Distance can be found on one of the following coordinates which is given below
(d,0), (0,d), (-d,0), (0,-d)
Ответ:
Explanation