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Two fixed, identical conducting plates (α &β ) , each of surface area S are charged to –Q and q, respectively, where Q > q > 0. A third identical plate (γ ), free to move is located on the other side of the plate with charge q at a distance d (Fig 1.13). The third plate is released and collides with the plate β . Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst β &γ . 

(a) Find the electric field acting on the plate γ before collision.  

(b) Find the charges on β and γ after the collision. 

(c) Find the velocity of the plate γ after the collision and at a distance d from the plate β.

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(b) During the collision plates, α & β are together and hence must be at one potential. Suppose the charge on β is q1 and on γ is q2. Consider a point O. The electric field here must be zero.

Thus the charge on β and γ are Q + q/2 and q/2, respectively.

c) Let the velocity be v at the distance d after the collision. If m is the mass of the plate γ, then the gain in K.E. over the round trip must be equal to the work done by the electric field. After the collision, the electric field at γ is

The work done when the plate γ is released until the collision is F1d where F1 is the force on plate γ.

The work done after the collision till it reaches d is F2d where F2 is the force on plate γ.

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