Question

A charged particle $(q, m)$ at rest is placed in an electric field $\vec{E}$ which varies with time as, $\vec{E} = E_0 \sin \frac{\omega_0 t}{2} \hat{i}$. The maximum velocity it can attain is

• A. $\frac{4qE_0}{m\omega_0}$
• B. $\frac{qE_0^2}{m\omega_0^2}$
• C. $\frac{2qE_0}{m\omega_0}$
• D. $\frac{2qE_0}{m\omega_0^2}$

Answer 1

Answer: (A)

$\frac{4qE_0}{m\omega_0}$

Answer 2

The force on the particle is

$$F = qE_0 \sin\left(\frac{\omega_0 t}{2}\right),$$

so the acceleration is

$$a = \frac{qE_0}{m} \sin\left(\frac{\omega_0 t}{2}\right).$$

Since the particle starts from rest, its velocity at time $t$ is

$$v(t) = \int_0^t a\,dt' = \frac{qE_0}{m} \int_0^t \sin\left(\frac{\omega_0 t'}{2}\right)dt'.$$

Let $u = \frac{\omega_0 t'}{2}$ so that $dt' = \frac{2}{\omega_0} du$. The limits change from $u=0$ to $u=\frac{\omega_0 t}{2}$. Thus,

$$v(t) = \frac{qE_0}{m} \cdot \frac{2}{\omega_0} \int_0^{\frac{\omega_0 t}{2}} \sin(u)\,du.$$

Evaluating the integral,

$$\int_0^{\frac{\omega_0 t}{2}} \sin(u)du = 1 - \cos\left(\frac{\omega_0 t}{2}\right),$$

so

$$v(t) = \frac{2qE_0}{m\omega_0} \left[1 - \cos\left(\frac{\omega_0 t}{2}\right)\right].$$

The maximum velocity occurs when the term $\left[1 - \cos\left(\frac{\omega_0 t}{2}\right)\right]$ is maximum. Since $\cos$ takes a minimum of $-1$,

$$\max \left[1 - \cos\left(\frac{\omega_0 t}{2}\right)\right] = 1 - (-1) = 2.$$

Thus,

$$v_{\text{max}} = \frac{2qE_0}{m\omega_0} \times 2 = \frac{4qE_0}{m\omega_0}.$$