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A particle of charge q and mass m moves rectilinearly under the action of an electric field E = alpha - betax . 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 - betax) "" …..(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 cdot (dv)/(dx) = q/m (alpha - betax)` `:. v dv = q/m int_0^x (alpha -betax) dx` `:.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 = 2alpha//beta`) and `a_("max") = qalpha//m ` (from Eq. (1)) |
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