Respuesta :
Answer:
6.86 cm
Explanation:
Given:
- q = charge on the first bead on the table= [tex]2.9\ nC = 2.9\times 10^{-9}\ C[/tex]
- m = mass of bead on the table = [tex]3.0\ mg = 3.0\times 10^{-6}\ kg[/tex]
- Q = charge on the second bead = [tex]-5.3\ nC = -5.3\times10^{-9}\ C[/tex]
Assume:
- r = the closest distance between the centers of the beads
- F = electrostatic force of attraction between the two beads
- W = weight of the first bead
- g = acceleration due to gravity = 9.8\ m/s^2
- N = normal force on the first bead
When the first bead rests on the table, then electrostatic force due to the second bead acts on it in the upward direction, Normal force acts in the upward direction and its weight in the downward direction.
So, using Newton's second law on the first bead resting on the table, we have
[tex]F+N-W=0\\[/tex]
At the closest distance of the second bead to the first bead, it just lifts off the table and the normal force becomes zero.
[tex]\therefore F-W=0\\\Rightarrow F=W\\\Rightarrow \dfrac{kqQ}{r^2}=mg\\\Rightarrow r^2=\dfrac{kqQ}{mg}\\\Rightarrow r^2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}\times 5.3\times 10^{-9}}{3\times 10^{-6}\times 9.8}\\\Rightarrow r^2=4.70\times 10^{-3}\\\textrm{Taking square root on both the sides}\\r = \pm 0.0686\ m\\\textrm{Since the distance is never negative}\\\therefore r = 0.0686\ m\\\Rightarrow r = 6.86\ cm[/tex]
Hence, the centers of the two beads must be brought closest to 6.86 cm before the lower bead is lifted off the table.
