# Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of the line AB.(a) If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it.(b) Assuming x << d, show that this force is proportional to x.(c) Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement?Find the time period of the oscillations if these conditions are satisfied.

Given:
Charge of particles A and B : q = Q
Separation between A and B : d
(a) here ‘x’ is the distance at which particle C of mass m and charge ‘q’ is place on the perpendicular bisector of AB.
Formula used:
From the figure we can find sinθ : The horizontal components of force cancel each other.
Total vertical component of force is: F’= 2Fsinθ
We use Coulomb’s Law: Where F is the electrostatic force on b due to A, k is a constant .
k = = 9× 109 Nm2C-2 and r is the distance between the two charges.
Here r = Substituting we get,  F’ is the net electric force experience by particle C of charge q.
(b)
When x
d, Thus, x2 can be neglected.
Substituting we get,   Thus, force is proportional to x.
(c)
The condition for Simple harmonic Motion of a particle is: Here, m is the mass of the particle C, ω is the angular frequency.
Thus comparing two equations of F’, we get We know that Where t is the Time Period.    Hence time period when the particle is released after a small displacement under SHM is Rate this question :

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