Answer:
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!!!
The magnitude of second charge is [tex]9.83 \times 10^{-12} \;\rm C[/tex].
Given data:
The magnitude of positive charged particle is, [tex]Q_{1}=+45 \;\rm nC = +45 \times 10^{-9}\;\rm C[/tex].
The mass of second charged particle is, [tex]m = 6.5 \;\rm \mu g=6.5\times 10^{-9} \;\rm kg[/tex].
The floating distance is, d = 25 cm = 0.25 m.
The concept behind this problem is that the force of attraction/repulsion between the two charges is equal to the force of gravity on the second charge. Then,
Force of gravity = Force of attraction/repulsion
[tex]m \times g=\dfrac{k \times Q_{1} \times Q_{2}}{d^{2}} \\\\6.5 \times 10^{-9} \times 9.8=\dfrac{9 \times 10^{9} \times 45 \times 10^{-9} \times Q_{2}}{(0.25)^{2}} \\\\Q_{2}= 9.83 \times 10^{-12} \;\rm C[/tex]
Thus, we can conclude that the magnitude of second charge is [tex]9.83 \times 10^{-12} \;\rm C[/tex].
Learn more about the Coulomb's force of attraction/repulsion here:
https://brainly.com/question/1616890
