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 with amplitude `(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

Explanation

`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))