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Question

A charged particle having charge q and mass m accelerated by a potential difference V0 enters inside a parallel plate capacitor (parallel to length l of plates) at mid-point of its separation between plates at t=0. Potential difference across capacitor varies with time t as, V=adt, where a is constant and d is separation between plates. Angle of deviation as it comes out of plates, is π4. Value of ma2I42qV30 , will be

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Solution

Horizontal distance

l=Vxt,

Velocity of charge particle in x- direction

Vx=2qV0m

Acceleration of charge particle is given as

a=qEm

Velocity of charge particle in y- direction

Vy=qEmdt

=qm at dt

Vy=qmat22

Vy=qmal22V2x

=qaI22m(m2qV0)

Vy=al24V0

tanθ=VyVx=al24m2qV30

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