# Two charges -q each are fixed separated by distance 2d. A third charge q of mass m placed at the midpoint is displaced slightly by x( x<

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by arnavsahay803

2015-04-26T23:49:50+05:30

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case 1)  displacement horizontally:

If the charge q is displaced slightly to the right side along the line joining the two charges -q, then the force of attraction of the right side -q is more than the force of attraction of the -q on the left side.

So +q moves towards right side. As it moves closer to the -q (right side), the force increases.  Then it will accelerate more and collide with it.

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case 2)   displacement of +q along the perpendicular bisector of line joining two charges -q and -q.

Let the direction of x be towards the top upwards from the center.  Let the origin be at the center of two charges -q.  Let the charge q be displaced by x , where x << d.   The distance between the two charges is 2 d.  Mass of charge q  is m.

The distance between -q and +q = √(d² + x²)

Let K = 1/[ 4 πε ].

see the diagram.  Let F1 be the force due to -q on the right and F2 be the force due to -q on the left side.   They are equal in magnitude and directions are as shown.

magnitude = F1 = F2 = K q² / (d² + x²)

Components of F1 and F2 along the perpendicular bisector are:
F1 Sin Ф = K q²  x / (x²+d²)³/²

The components of these forces parallel to the line joining -q charges are = F1 Cos Ф and will cancel as they are in opposite directions.

Let F and a be the instantaneous resultant force and net acceleration of +q.

Then the resultant force on q due to the two charges:
F =  2 K q² x / (d²+x²)³/²  towards the origin in the direction of - x.

m d² x/ dt²   = - 2 K q² x / [d³ (1 + x²/d²)³/² ]
=  - 2 K q² x / d³   if  x << d, then we ignore x² as it is << d²

m  d² x / d t² =  - [ 2 q² / 4 π ε d³ ]  x
= ⁻ω² x
This is the equation of motion of the charge q.  SO q oscillates in a simple harmonic motion with an amplitude of  x₀ that is the displacement at t = 0.

The angular frequency of oscillation =  ω  = q / √ (2π ε d³)

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