Answer:
The fraction of the electrons from the first sphere that must be transferred to the second sphere is 2.2687 X 10¹⁶ Electrons
Explanation:
From coulomb's law;
[tex]F = \frac{K*q_1q_2}{r^2}[/tex]
where;
F is the electrostatic force between the two spheres = 200 X 10³ N
k is coulomb's constant = 8.99 X 10⁹ Nm²/C²
q₁ is the charge on first sphere
q₂ is the charge on second sphere
r is the distance between the two spheres = 77cm = 0.77m
Since the two sphere have equal mas, the magnitude of their charge will be same = q
[tex]F = \frac{K*q^2}{r^2}[/tex]
[tex]q^2 = \frac{F*r^2}{K}, then, q = \sqrt{\frac{F*r^2}{K}} = r\sqrt{\frac{F}{K}}[/tex]
[tex]q = (0.77)\sqrt{\frac{200000}{8.99X10^9}} = 0.0036344 C[/tex]
q = 3.6344 X 10⁻³C
The magnitude of charge on each sphere is 3.6344 X 10⁻³C
To find the number of electrons transferred
N = q/e
where;
N is the number of electrons transferred
q is the charge on each sphere
e is the charge on one electron = 1.602 X 10⁻¹⁹ C
N = (3.6344 X 10⁻³)/(1.602 X 10⁻¹⁹)
N = 2.2687 X 10¹⁶ Electrons
The fraction of the electrons from the first sphere that must be transferred to the second sphere is 2.2687 X 10¹⁶ Electrons
Answer:
78000 electrons
Explanation:
The force of attraction between the two charges is given as
F = Kq1q2/ r^2
Since the charges are equal
q1= q2
F = kq^2/r^2
Making q as the subject of the formulas
q = r × √F/k
q = 77× 10^-2 m × √ 200× 10^3 / 9 × 10^ 9 Nm^2/C^2
= 0.77 × √ 22.22× 10^-6
=0.77 × 4.714 × 10^ -3
q = 3.63 ×10^-3 C
= 3.63mC
The fraction of electron to be transferred from the first sphere to the second sphere
N = q/e ; relationship between number of electrons and charge
With copper atom having 29 electrons
= 3.63 × 10^-3/ 1.6 × 10^-9 ×29
3.63 × 10^-3 /46.4 × 10^ -9
N = 0.078 × 10 ^ 6
N =78000 electrons that will be transferred
