A positively charged bead of mass m and unknown charge (denoted q) moves with no

Question

A positively charged bead of mass m and unknown charge (denoted q) moves with no friction along a vertical rod, as sketched in Fig. 5, subject both to the Earth gravity and to the electric force due to another positive charge Q fixed at the lower end of the rod. The bead is observed to execute an oscillating motion between "levels" A and B shown in the figure along with the relevant geometrical dimensions. Based on this observation, find charge q of the bead. The bead and fixed charge have small sizes and can be treated as point-like objects.

Explanation / Answer

Forces on the bead at position A are gravity and the electrostatic force.
gravity force will be downward while the electrostatic force will be upward .
hence at point A the two forces must be same and when it try to move down below it will
experrience an upward force.
Hence at point A
GMm /r2 = kqQ /h2 ------------(1)
And Similarly at the point B
GMm /(r+d)2 = kqQ /(h+d)2 -----------(2)
where G is gravitational constant , M is mass of earth and r is the distance of the bead from earth centre.
k is coloumb's constant
From equation 1
GMm = r2(kqQ/h2)
putting this value in eqaution 2
r2 /(r+d)2 = h2 /(h+d)2
taking square root and reciprocal of them
(r+d)/r = (h+d)/h
1+d/r = 1+d/h
hence r = h
Now putting this inn equation 1
GMm = kqQ
q = GMm /kQ
putting the value of constant
q = (6.67*10-11)*(5.972*1024)*m /(9*109)Q
q = 4.427*104*(m/Q)

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