Answer:
[tex]1.04461\times 10^{-9}\ C[/tex]
Explanation:
Charge of particle = q
The particles have same charge
r = Distance between charge
k = Coulomb constant = [tex]8.99\times 10^9\ Nm^2/C^2[/tex]
m = Mass of particle = 10 mg
g = Acceleration due to gravity = 9.81 m/s²
Electrostatic force is given by
[tex]F=k\frac{q_1q_2}{r^2}[/tex]
According to Newton's second law
[tex]F=mg[/tex]
Here, both the electrostatic and inertial force is conserved
[tex]mg=k\frac{q^2}{r^2}\\\Rightarrow q=\sqrt{\frac{mgr^2}{k}}\\\Rightarrow q=\sqrt{\frac{10\times 10^{-6}\times 9.81\times 0.01^2}{8.99\times 10^9}}\\\Rightarrow q=1.04461\times 10^{-9}\ C[/tex]
The charge of the particles is [tex]1.04461\times 10^{-9}\ C[/tex]
The charge of the particles is very low as compared to typical static electricity
(a) The magnitude of the charge is obtained as [tex]1.04\times 10^{-9}\,C[/tex]
(b) The magnitude of this charge is much less than what is typical of static electricity.
(a) The pieces have equal charge 'q'.
Distance between the charges, [tex]r=1\,cm=0.01\,m[/tex]
Given that the electrostatic repulsive force balances the weight of the top piece. So,
[tex]k\frac{q^2}{r^2} =mg[/tex]
Where, [tex]k = 9\times 10^9 \,Nm^2/C^2[/tex] is the Coulomb's constant.
Substituting the known values, we get;
[tex](9\times 10^9\, Nm^2/C^2)\frac{q^2}{(0.01\,m)^2} =10\times 10^{-6}kg \times 9.8\,m/s^2[/tex]
[tex]\implies q=\sqrt{\frac{10\times 10^{-6}\,kg \times 9.8\,m/s^2 \times (0.01)^2}{9\times 10^9 Nm^2/C^2} } = \sqrt{1.088\times 10^{-18}\,C^2} =1.04\times 10^{-9}\,C[/tex]
(b) This charge is much less when compared with a typical charge in the case of static electricity.
Learn more about Coulomb's law here; https://brainly.com/question/14049417
