Consider a particle of mass 1 text{ g} and ch | vedclass

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(B) The force on the particle is $F = qE = qE_0 \sin(\omega t)$.Using Newton's second law, $ma = qE_0 \sin(\omega t)$, so $a = \frac{qE_0}{m} \sin(\om

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$4$

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(B) The force on the particle is $F = qE = qE_0 \sin(\omega t)$.Using Newton's second law, $ma = qE_0 \sin(\omega t)$, so $a = \frac{qE_0}{m} \sin(\omega t)$.The velocity $v(t)$ is the integral of acceleration: $v(t) = \int a \, dt = \int \frac{qE_0}{m} \sin(\omega t) \, dt = -\frac{qE_0}{m\omega} \cos(\omega t) + C$.Since the particle starts from rest at $t = 0$, $v(0) = 0$, which gives $C = \frac{qE_0}{m\omega}$.Thus, $v(t) = \frac{qE_0}{m\omega} (1 - \cos(\omega t))$.The maximum speed occurs when $\cos(\omega t) = -1$, so $v_{\max} = \frac{2qE_0}{m\omega}$.Substituting the values: $q = 1 	ext{ C}$, $E_0 = 2 	ext{ N/C}$, $m = 1 	ext{ g} = 10^{-3} 	ext{ kg}$, and $\omega = 1000 	ext{ rad/s}$.$v_{\max} = \frac{2 	imes 1 	imes 2}{10^{-3} 	imes 1000} = \frac{4}{1} = 4 	ext{ m/s}$.

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