Maximum Speed of Charged Particle in Time Varying Sinusoidal Electric Field

To find the maximum speed attained by the charged particle, we start by applying Newton's second law of motion to describe the particle's movement under the influence of the given electric field.

The electrostatic force $\vec{F}(t)$ acting on the particle with charge $q$ in an electric field $\vec{E}(t) = E_0 \sin(\omega t) \hat{i}$ is given by: $\vec{F}(t) = q \vec{E}(t) = q E_0 \sin(\omega t) \hat{i}$

Using Newton's second law, the acceleration $\vec{a}(t)$ of the particle of mass $m$ along the x-axis is: $a(t) = \frac{dv(t)}{dt} = \frac{q E_0}{m} \sin(\omega t)$

Since the particle is initially at rest at $t = 0$ (i.e., $v(0) = 0$), we can find its velocity $v(t)$ at any subsequent time $t$ by integrating the acceleration with respect to time: $v(t) = \int_0^t a(t') \, dt' = \int_0^t \frac{q E_0}{m} \sin(\omega t') \, dt'$

Executing the integration: $v(t) = \frac{q E_0}{m} \left[ -\frac{\cos(\omega t')}{\omega} \right]_0^t = \frac{q E_0}{m \omega} (1 - \cos(\omega t))$

The term $(1 - \cos(\omega t))$ oscillates between $0$ and $2$. Therefore, the velocity $v(t)$ reaches its maximum value $v_{\text{max}}$ when $\cos(\omega t) = -1$: $v_{\text{max}} = \frac{2 q E_0}{m \omega}$

Now, substituting the given numerical values:

• $q = 1.0\text{ C}$
• $E_0 = 1.0\text{ N C}^{-1}$
• $m = 10^{-3}\text{ kg}$
• $\omega = 10^3\text{ rad s}^{-1}$

$v_{\text{max}} = \frac{2 \times 1.0 \times 1.0}{10^{-3} \times 10^3} = \frac{2}{1} = 2\text{ m s}^{-1}$

Thus, the maximum speed attained by the particle is $2$ (or $2.00$).