Question

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 700-mm strings attached to the same nail in a ceiling. Surplus electrons are added to each sphere, and then the spheres are brought in contact with each other and released. Their equilibrium position is such that each string makes a 17.0° angle with the vertical.
How many surplus electrons are on each sphere?

Answer 1

Answer:

The number of electrons surplussed on each surface is $\bf{2 \times 10^{21}}$.

Explanation:

Given:

The masses of the sphere, $m = 9.10~g$

The length of the strings, $L = 700~mm$

The angle made by each string with vertical, $\theta = 17.0^{0}$

According to the diagram, the equilibrium condition for the vertical components of the forces acting on each sphere can be written as

$T \cos \theta = mg~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

The equilibrium condition for the horizontal components of the forces acting on each sphere can be written as

$T \sin \theta = F~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Here, $F$ is the electrostatic force experienced by the metal spheres.

The value of the electrostatic force is given by

$F = \dfrac{q^{2}}{r^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Here, $q$ is the charge on each sphere and $r$ is the distance between them.

Referring to the figure, from $\bigtriangleup QOR$,

$&& \dfrac{r/2}{L} = \sin \theta\\&or,& r = 2L \sin \theta$

Substituting the value of $r$ in equation (3), we have

$F = \dfrac{q^{2}}{(2L \sin \theta)^{2}}~~~~~~~~~~~~~~(4)$

From equation (1),

$T &=& \dfrac{mg}{\cos \theta}\\&=& \dfrac{(9.10~g)(980~cm/s^{2})}{\cos 17^{0}}\\&=& 9325.7~dyn$

Substituting the values of $T$ and $F$ in equation (2), we have

$&& (9325.7~dyn) \sin 17^{0} = \dfrac{q^{2}}{(2 \times 70 \times \sin 17^{0})}\\&or,& q^{2} = ((9325.7~dyn) \sin 17^{0})(2 \times 70 \times \sin 17^{0})\\&or,& q = 334~C$

If $n$ is the number if electron surplussed with the metal sphere, then

$ne = q~~~~~~~~~~~~~~~~~~~~~~~~~~(5)$

Here, $e$ is the electronic charge.

Substituting the value of $e$ and $q$ in equation (5), we have

$n &=& \dfrac{q}{e}\\~~~&=& \dfrac{334~C}{1.6 \times 10^{-19}~C}\\~~~&=& 2 \times 10^{21}$