Question

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,

• A. The motion of the particle is oscillatory
• B. The amplitude of the particle is $\left(\frac{\alpha }{\beta }\right)$
• C. The mean position of the particle is at $x=\left(\frac{\alpha }{\beta }\right)$
• D. The maximum acceleration of the particle is $\frac{q\alpha }{m}$

Answer 1

Answer: (A), (B), (C), (D)

The maximum acceleration of the particle is $\frac{q\alpha }{m}$

$F=qE$
$ma=q(\alpha -\beta x)$
$a=\frac{q}{m}(\alpha -\beta x$)

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

At mean position, we know that $F=0$
$⟹a=0$
$⟹\frac{q}{m}(\alpha -\beta x)$ = 0
$⟹(\alpha -\beta x)$ = 0
$⟹x_m=\frac{\alpha }{\beta }$

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

$a=\frac{q}{m}(\alpha -\beta x$)
$⟹v\frac{dv}{dx}=\frac{q}{m}(\alpha -\beta x)$
$⟹\int vdv=\int \frac{q}{m}(\alpha -\beta x)dx$
$⟹\frac{v^2}{2}=\frac{q}{m}(\alpha x-\beta \frac{x^2}{2})$
As $v=0$
We get $x=0$ and $x=\frac{2\alpha }{\beta }$

$\therefore$ Amplitude of the particle $=x-x_m$ $=\frac{2\alpha }{\beta }-\frac{\alpha }{\beta }=\frac{\alpha }{\beta }$

Maximum acceleration of the particle = $a_m=\frac{q}{m}(\alpha -\beta \times \frac{2\alpha }{\beta }$)
$\therefore a_m=\frac{q}{m}\left(\alpha \right)$ (ignore the -ve sign)