Two neutral insulating small spheres are rubbed against each other and...
Solution:
Charge on each sphere:
Let Q1 and Q2 be the charges on the two spheres. The force of attraction between the two spheres can be given by Coulomb's law as:
F = k * Q1 * Q2 / r^2
where k is Coulomb's constant, r is the separation between the spheres.
In this case, the spheres are neutral before rubbing, so Q1 = Q2 = 0. After rubbing, electrons get transferred from one sphere to the other, resulting in a net charge on each sphere.
Let Q be the charge on each sphere after rubbing. The force of attraction can now be written as:
F = k * Q^2 / r^2 = 3.6 N
Solving for Q, we get:
Q = sqrt(3.6 N * r^2 / k) = 1.2 x 10^-8 C
Therefore, the charge on each sphere is 1.2 x 10^-8 C.
Number of electrons transferred:
The charge on one electron is -1.6 x 10^-19 C. To find the number of electrons transferred, we can divide the total charge on one sphere by the charge on one electron:
N = Q / (-1.6 x 10^-19 C) = 7.5 x 10^10 electrons
Therefore, 7.5 x 10^10 electrons get transferred from one sphere to the other during rubbing.
Explanation:
When two insulating small spheres are rubbed against each other, electrons get transferred from one sphere to the other due to friction. This results in a net charge on each sphere, and they attract each other due to the electrostatic force of attraction.
The force of attraction between the two spheres can be calculated using Coulomb's law, which relates the force to the charges on the spheres and the separation between them. Solving for the charge on each sphere, we get 1.2 x 10^-8 C.
To find the number of electrons transferred, we divide the total charge on one sphere by the charge on one electron. This gives us 7.5 x 10^10 electrons, which is the number of electrons transferred from one sphere to the other during rubbing.
Two neutral insulating small spheres are rubbed against each other and...