Two charges ,each equal to q, are kept at x=-a and x=a on the x-axis.A particle of mass m and charge q0=q/2 is placed at origin .If charge q0 is given by a small displacement y(y< Answer is y

Plz solve from easy method

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Then we have. To find equilibrium position ?

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2015-04-03T03:08:39+05:30

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Let the charge q₀ be given a small displacement y towards the positive x axis.  Let the instantaneous position of q₀ be x(t) and x(0) = y.  Also, the initial velocity of q₀ is 0.

Let K = (1/4πε₀)
The net force acting on q₀ will be in the negative x direction, as the force exerted by the charge q at x = a exerts higher force.

Net Force acting on q₀ =  - K q q₀ [ 1/(a-x)² - 1/(a+x)² ]
         F = m d² x(t)/ dt² = - K q q₀ [ 4 a x  /(a² - x²)² ]
           d² x(t) / dt² = - 4 a k q q₀/m  x / [ a⁴ * (1 - x²/a²)² ]
                       Let ω²  = 4 k q q₀ / (m d³)  = q q₀ /(π ε₀ m d³)

for  x , y << a,

     d² x(t) / dt² ≈ - ω² x [ 1 + 2 x²/a² ]  ≈  -ω² x(t)

This is a simple harmonic motion. SHM
    The angular frequency of oscillation = ω = √[ q q₀ / (π ε₀ m d³) ]
     T = time period = 2π/ω = 2π √[ (π ε₀ m d³) / (q q₀) ]

 the instantaneous displacement is :  x(t) = y Cos (ω t)
           as  x = y , at  t = 0 sec.

The amplitude of oscillation is  y.

we can calculate the velocity, acceleration, and energy in the oscillations.

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