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

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)

a, c, d, e are all applying about success strategies.

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