A particle,of mass 10^{-3} kg and charge 1.0 | vedclass

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Question

(A) The force on the particle is given by $F = qE = qE_0 \sin(\omega t)$.Using Newton's second law,$F = ma$,we have $a = \frac{qE_0}{m} \sin(\omega t)

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The force on the particle is given by $F = qE = qE_0 \sin(\omega t)$.Using Newton's second law,$F = ma$,we have $a = \frac{qE_0}{m} \sin(\omega t)$.Since $a = \frac{dv}{dt}$,we integrate with respect to time:$v(t) = \int_0^t \frac{qE_0}{m} \sin(\omega t') dt' = \frac{qE_0}{m\omega} [-\cos(\omega t')]_0^t = \frac{qE_0}{m\omega} (1 - \cos(\omega t))$.The speed is maximum when $\cos(\omega t) = -1$,which occurs at $\omega t = \pi, 3\pi, \dots$.At these times,$v_{\max} = \frac{qE_0}{m\omega} (1 - (-1)) = \frac{2qE_0}{m\omega}$.Substituting the given values: $q = 1.0 \ C$,$E_0 = 1.0 \ N \ C^{-1}$,$m = 10^{-3} \ kg$,and $\omega = 10^3 \ rad \ s^{-1}$.$v_{\max} = \frac{2 	imes 1.0 	imes 1.0}{10^{-3} 	imes 10^3} = \frac{2}{1} = 2 \ m \ s^{-1}$.

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