What is the number of excess electrons on each sphere in the following case?
Two small spheres spaced 35 cm apart have an equal charge. How many excess electrons must be present on each sphere so that the magnitude of the force of repulsion between them is 2.20*10^21 N.
The force between two bodies with a charge q1 and q2 and a distance of separation r is given by Coulomb's law as q1*q2/(4*pi*e0*r^2). 1/(4*pi*e0) = 8.987*10^9 Nm^2/C^2
Now the distance between the spheres in the problem is 35 cm or 0.35 m.
The force of repulsion is supposed to be 2.20*10^21 N.
Now the charge on an electron is −1.602* 10^-19 C
If there are n electrons on each sphere:
n^2* (-1.602*10^-19)^2*8.987*10^9/0.35 = 2.20*10^21
=> n^2 = 2.2* 10^21*.35 / (-1.602*10^-19)^2*8.987*10^9
=> n^2 = 3.3*10^48
So n can be taken to be approximately 1.8*10^24.
Therefore each sphere should have an excess of 1.8*10^24 electrons.
We'll write the Coulomb force formula:
F = (q^2/r^2)*K, where k is electrostatic constant; k = 8.988*10^9Nm^2C^-2.
We know the force F and the the distance between spheres:
F = 2.20*10^21N and r = 0.35m (conversion from cm to m)
F*r^2*K = q^2
q = sqrt (F*r^2*K)
q = sqrt (2.20*10^21*0.1225*8.988*10^9)
q = 0.17*10^-15 C
The charge of a single electron is e = 1.6*10^-19 C, so the number of excess electrons is:
N = q/e
N = 0.17*10^-15 C/1.6*10^-19 C
N = 1062 approx.