Respuesta :
As the each sphere have the same charge, we use the Coulomb law to calculate the charge on the sphere
[tex]F=k\frac{qq}{r^2} \\\\q=\sqrt{\frac{F r^2}{k} }[/tex]
Here, [tex]F=3.33\times 10^{-21} N[/tex] and [tex]r=20 cm=20\times10^{-2} m[/tex] and k is constant and its value is  [tex]8.99\times 10^{9} N.m^2/C^2[/tex]
Therefore,
[tex]q=\sqrt{\frac{3.33\times10^{-21}\times(20\times10^{-2} m)^2 }{8.99\times 10^{9} N.m^2/C^2 } }\\\\q=1.2\times 10^{-16} C[/tex]
Thus, the number of electron present in each sphere is
[tex]N=\frac{q}{e} \\\\ N =\frac{1.2\times 10^{-16} C}{1.6\times10^{-19}C } \\\\ N=0.75\times 10^3=750[/tex]
Solution:
Since we have two small spheres:
One Charge = q1 = q Â
Force = F = 4.57*10^-21 N Â
Other charge = q2 =q Â
Distance = r = 20 cm = 0.2 m Â
permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2) Â
Using Coulomb's law, Â
F=[1/4pieo]q1q2/r^2 Â
F = [1/4pieo]q^2 / r^2 Â
q^2 =F [4pieo]r^2 Â
q = r*sq rt F[4pieo] Â
q=0.2* sq rt[3.33 x 10^-21]*[4*3.1416*8.854*10^-12] Â
q = 1.28923*10^ -16 C Â
So,
number of excess electrons = n = q/e=1.28923*10^ -16 /1.6*10^-19 Â
n =750 Â
Thus,
750 excess electrons must be present on each sphere.
