Given, mass of particle = 10^$$-$$3 kg, charge on particle = 1.0 C, electric field $$\overrightarrow E (t) = {E_0}\sin \omega t\,\widehat i$$, E₀ = 1.0 N C^$$-$$1, $$\omega$$ = 10³ rad s^$$-$$1

Force on particle is given by

$$ \Rightarrow \overrightarrow F = q{E_0}\sin \omega t\,\widehat i = 1.0 \times 1.0 \times \sin ({10^3}t)\widehat i$$

$$ \Rightarrow \overrightarrow F = \sin ({10^3}t)\widehat i$$

We know that $$\overrightarrow F = m\overrightarrow a \Rightarrow \overrightarrow a = {{\overrightarrow F } \over m}$$

$$ \Rightarrow a = {{\sin ({{10}^3}t)} \over {{{10}^{ - 3}}}} \Rightarrow a = {10^3}\sin ({10^3}t)$$

We know $$a = {{dv} \over {dt}}$$

$$ \Rightarrow {{dv} \over {dt}} = {10^3}\sin ({10^3}t)$$

$$ \Rightarrow dv = {10^3}\sin ({10^3}t)dt$$

Now, integrating it, we get

$$\int\limits_0^v {dv = \int\limits_0^t {{{10}^3}\sin ({{10}^3}t)dt} } $$

$$ \Rightarrow \left. v \right|_0^v = {10^3}\left( {{{ - \left. {\cos ({{10}^3}t)} \right|_0^t} \over {{{10}^3}}}} \right) \Rightarrow \left. v \right|_0^v = \left. { - \cos ({{10}^3}t)} \right|_0^t$$

$$ \Rightarrow v = ( - \cos {10^3}t + \cos 0)$$

$$ \Rightarrow v = (1 - \cos ({10^3}t))$$

We know that cos$$\theta$$ can take values between $$-$$1 and 1. Therefore, maximum speed attained by the particle when cos$$\theta$$ = $$-$$1. Thus,

$${v_{\max }} = 1 - ( - 1) = 2$$ m s^$$-$$1