Answer:
The number of electrons surplussed on each surface is [tex]\bf{2 \times 10^{21}}[/tex].
Explanation:
Given:
The masses of the sphere, [tex]m = 9.10~g[/tex]
The length of the strings, [tex]L = 700~mm[/tex]
The angle made by each string with vertical, [tex]\theta = 17.0^{0}[/tex]
According to the diagram, the equilibrium condition for the vertical components of the forces acting on each sphere can be written as
[tex]T \cos \theta = mg~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
The equilibrium condition for the horizontal components of the forces acting on each sphere can be written as
[tex]T \sin \theta = F~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Here, [tex]F[/tex] is the electrostatic force experienced by the metal spheres.
The value of the electrostatic force is given by
[tex]F = \dfrac{q^{2}}{r^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]
Here, [tex]q[/tex] is the charge on each sphere and [tex]r[/tex] is the distance between them.
Referring to the figure, from [tex]\bigtriangleup QOR[/tex],
[tex]&& \dfrac{r/2}{L} = \sin \theta\\&or,& r = 2L \sin \theta[/tex]
Substituting the value of [tex]r[/tex] in equation (3), we have
[tex]F = \dfrac{q^{2}}{(2L \sin \theta)^{2}}~~~~~~~~~~~~~~(4)[/tex]
From equation (1),
[tex]T &=& \dfrac{mg}{\cos \theta}\\&=& \dfrac{(9.10~g)(980~cm/s^{2})}{\cos 17^{0}}\\&=& 9325.7~dyn[/tex]
Substituting the values of [tex]T[/tex] and [tex]F[/tex] in equation (2), we have
[tex]&& (9325.7~dyn) \sin 17^{0} = \dfrac{q^{2}}{(2 \times 70 \times \sin 17^{0})}\\&or,& q^{2} = ((9325.7~dyn) \sin 17^{0})(2 \times 70 \times \sin 17^{0})\\&or,& q = 334~C[/tex]
If [tex]n[/tex] is the number if electron surplussed with the metal sphere, then
[tex]ne = q~~~~~~~~~~~~~~~~~~~~~~~~~~(5)[/tex]
Here, [tex]e[/tex] is the electronic charge.
Substituting the value of [tex]e[/tex] and [tex]q[/tex] in equation (5), we have
[tex]n &=& \dfrac{q}{e}\\~~~&=& \dfrac{334~C}{1.6 \times 10^{-19}~C}\\~~~&=& 2 \times 10^{21}[/tex]
