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Question

A particle of charge q and mass m moves rectilinearly under the action of an electric field E=αβx. Here, α and β are positive constants and x is the distance from the point where the particle was initially at rest. Then,

A
The motion of the particle is oscillatory
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B
The amplitude of the particle is (αβ)
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C
The mean position of the particle is at x=(αβ)
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D
The maximum acceleration of the particle is qαm
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Solution

The correct option is D The maximum acceleration of the particle is qαm
F=qE
ma=q(αβx)
a=qm(αβx)

As a depends upon x, it means it is certainly an oscillatory motion.

At mean position, we know that F=0
a=0
qm(αβx) = 0
(αβx) = 0
xm=αβ

In order to find the amplitude, we must remember that when the particles reaches its amplitude, velocity becomes 0.

a=qm(αβx)
vdvdx=qm(αβx)
vdv=qm(αβx)dx
v22=qm(αxβx22)
As v=0
We get x=0 and x=2αβ

Amplitude of the particle =xxm =2αβαβ=αβ

Maximum acceleration of the particle = am=qm(αβ×2αβ)
am=qm(α) (ignore the -ve sign)

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