A charged particle `(q,m)` is released from origin in an electric field `E = alpha - beta x` where `alpha` and `beta` are constants and `x` is distance from origin. Find the velocity of particle in terms of `x` and maximum displacement of the particle.

Explanation

When the particle is at distance `x` from origin , the force on it is `F = qE`.
Acceleration of a particle , `a = (qE)/(m) = (q(alpha - beta x))/(m)`
Here , acceleration `a` is a variable .
`v(dv)/(dx) = (q(alpha - beta x))/(m)`
`int_(0)^(v) vdx = (q)/(m) int_(0)^(x) (alpha - beta x)dx`
`|(v^(2))/(2)|_(0)^(v) = (q)/(m) |alpha x - (beta x^(2))/(2)|_(0)^(x)`
`(v^(2))/(2) = (q)/(m) (alpha x - (beta x^(2))/(2))`
`v = sqrt((q)/(m) (2 alpha x - beta x^(2)))`
when particle stops , `v = 0` .
`x = x_(0) = ( 2 alpha)/(beta)`