Q. 323.9( 7 Votes )

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.

Answer :

Charge of particles A and B : q = Q
Separation between A and B : d

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.
When x
Thus, x2 can be neglected.
Substituting we get,

Thus, force is proportional to x.
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

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