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A particle of charge q and mass m moves rectilinearly under the action of an electric field E= alpha- beta x. Here, alpha and beta are positive constants and x is the distance from the point where the particle was initially at rest. Then: (1) the motion of the particle is oscillatory withamplitude (alpha)/(beta) (2) the mean position of the particles is at x= (alpha)/(beta) (3) the maximum acceleration of the particle is (q alpha)/(m) (4) All 1, 2 and 3 |
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Answer» SOLUTION :`a= (F)/(m) = (qE)/(m) = (q)/(m) (alpha - beta X)` …(1) a= 0 at `x = (alpha)/(beta)` i.e., force on the particle is zero at `x= (alpha)/(beta)`, So, mean POSITION of particle is at `x= (alpha)/(beta)` Eq (1) can be WRITTEN as `v.(dv)/(dx) = (q)/(m) (alpha - beta x)` `therefore v dv = (q)/(m) int_(0)^(x) (alpha - beta x) dx` `therefore v= sqrt((2qx)/(m)(alpha-(beta)/(2)x))` v=0 at x = 0 and `x = (2 alpha)/(beta)` so, the particle will oscillate between x = 0 to `x= (2 alpha)/(beta)` with mean position at `x =(alpha)/(beta)`. Therefore, amplitude of particle is `(alpha)/(beta)`. Maximum acceleration of particle is at extreme positions (at x= 0 or x= `2 alpha//beta`) and `a_("max") = q alpha//m` (from Eq.(1)) |
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