Question

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

Answer 1

Answer: 2

$m=10^{-3}kg,q=1C,t=0$
$E=E_0\sin \omega t=$
Force on particle will be
$F=qE=q{E}_0\sin \omega t$
At $v_{\max },F=0\Rightarrow q{E}_0\sin w{t}_0=0$
$\Rightarrow \omega t_0=\pi$
$F=q{E}_0\sin \omega t$
$\underset{0}{\overset{v}{\int }}dv=\underset{0}{\overset{\pi /\omega }{\int }}\frac{q{E}_0}{m}\sin \omega dt$
$v-0=\frac{q{E}_0}{m\omega }[-\cos \omega t{]}_0^{\pi /\omega }$
$v-0=\frac{q{E}_0}{m\omega }\lfloor (-\cos \omega )-(-\cos 0)]$
$v=\frac{1\times 1}{m\omega }\times 2=2m/s=2.00$