Question

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

Answer 1

$E\left(t\right)=E_{0}sin(\omega t$)
$F_E=qE\left(t\right)$
$\frac{dv}{dt}=\frac{qE\left(t\right)}{m}$
$⟹{\int }_0^{v_{max}}dv=\frac{q{E}_{0}sin\left(\omega t\right)}{m}dt$
$⟹v_{max}=\frac{q{E}_0}{\omega m}{\int }_0^t-cos\omega t$
$⟹v_{max}=\frac{q{E}_0}{\omega m}[1-cos(\omega t\left)\right]$
For $v_{max}$, we need to have value of
$\cos \omega t$ = -1
$\therefore v_{max}=\frac{2q{E}_0}{\omega m}=2m/s$