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, the

Options

A. motion of the particle is oscillatory
B. amplitude of the particle is `(alpha//beta)`
C. mean position of the particle is at `x=(alpha//beta)`
D. maximum acceleration of the particle is `(qalpha)/m`.

Answer

A, B, C, D

Explanation

Acceleration of particle,
`a=F/m = (qE)/m=q/m(alpha-betax)` ….(i)
a=0 at x=`alpha/beta`
i.e., force on the particle is zero at x= `alpha/beta`
So, mean position of the particle is at x= `alpha/beta`
`v(dv)/(dx)=q/m(alpha-betax)`
`therefore int_0^v vdv=q/mint_0^x (alpha-betax)dx " " therefore v=sqrt((2qx)/m(alpha-beta/2x))`
v=0 at x=0 and `x=(2alpha)/beta`
So, the particle oscillates between x=0 and `x=(2alpha)/beta` with mean position at `x=alpha/beta`
Maximum acceleration of the particle is at extreme position (at x=0 or `x=2alpha//beta` and `a_"max"=qalpha//m` [ from equation (i)]