A positively charged particle Q1 = +45 nC is held fixed at the origin. A second charge Q2 of mass m = 6.5 μg is floating a distance d = 25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth.

Q₂ = (9.83 × 10⁻⁹) C = +9.83 nC

Explanation:

The force of attraction/repulsion between the two charges is equal to the force of gravity on the charge 2.

The force of gravity on the charge two = mg

= 6.5 μg × 9.8 m/s² = (6.37 × 10⁻⁵) N

Since the gravity force is directed downwards, the force on charge 2 due to charge 1 has to be directed upwards, hence it is a force of repulsion.

The magnitude of the force between the two charges is given according to the Coulomb's law.

F = kQ₁Q₂/r²

k = Coulomb's constant = (9.0 × 10⁹) Nm²/C²

Q₁ = 45 nC = (45 × 10⁻⁹) C

Q₂ = ?

r = 25 cm = 0.25 m

F = (6.37 × 10⁻⁵) N

(6.37 × 10⁻⁵) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] ÷ 0.25²

Q₂ = 0.00000000983 C = (9.83 × 10⁻⁹) C = 9.83 nC

Hope this Helps!!!

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form