Physics - Electrostatics Question with Solution | TestHub

$\vec{E}=(10-5x)i$ $\mathrm{m}=5\mathrm{kg},\overrightarrow{\mathrm{F}}=\mathrm{q}\overrightarrow{\mathrm{E}}$ $\overrightarrow{\mathrm{F}}=(10-5\mathrm{x})\mathrm{i}$ (a) F $\propto -\mathrm{x}$ So motion of charge particle is oscillatory (b) F=0 at $\mathrm{x}=2$ oscillation about $\mathrm{x}=2$ so maximum displacement is 4 $\mathrm{F}=10-5\mathrm{x}$ $\mathrm{a}=\frac{10-5\mathrm{x}}{2}$ $\mathrm{a}=2-\mathrm{x},\mathrm{a}=-(\mathrm{x}-2)$ so ${\omega }^2=1,\omega =1$ (c) ${\mathrm{v}}_{\text{max }}=\mathrm{A}\omega$ $\mathrm{A}=2,\omega =1$ ${\mathrm{v}}_{\text{max }}=2$ (d) velocity at mean position at $\mathrm{x}=2$