Question

A particle, of mass ${10}^{–3}kg$ and charge $1.0C,$ is initially at rest. At time $t=0sec$ the particle comes under the influence of an electric field $\overrightarrow{E}\left(t\right)=E_0\sin \omega t\hat{i},$ where $E_0=1.0\text{N}{\text{C}}^{-1}$ and $\omega ={10}^3$ rad $s^{–1}$. Consider the effect of only the electrical force on the particle. Then the maximum speed, in ${\text{m s}}^{–1,}$ attained by the particle at subsequent times is

Answer 1

Given,
$m={10}^{-3}kg,q=1.0C\overrightarrow{E}\left(t\right)=E_0\sin \omega t\hat{i}$
and as we know,
change in liner moment = impulse
so, $m\frac{dv}{dt}=qE$
taking integration at both side
$\Rightarrow {\int }_0^{v}dv=\frac{q{E}_0}{m}{\int }_0^t\sin \omega tdt$
$v=\frac{q{E}_0}{m}\frac{(1-\cos \omega t)}{\omega }=\frac{1\times 1(1-\cos \omega t)}{{10}^{-3}\times {10}^3}$
$v=(1-\cos \omega t)$
$\therefore V_{max}=2m/s$