Step 1: Force on the Particle
The force on the particle due to the electric field is given by:
\[
F = q E(t)
\]
Where:
- \( q = 1.0 \, \text{C} \) (charge of the particle),
- \( E(t) = E_0 \sin(\omega t) \) (electric field).
Thus:
\[
F = 1.0 \cdot 2 \sin(1000 t) = 2 \sin(1000 t) \, \text{N}
\]
Step 2: Acceleration of the Particle
The acceleration of the particle is given by Newton's second law:
\[
F = m a
\]
Where:
- \( m = 1 \, \text{gm} = 1 \times 10^{-3} \, \text{kg} \) (mass of the particle),
- \( a \) is the acceleration of the particle.
Thus:
\[
a = \frac{F}{m} = \frac{2 \sin(1000 t)}{1 \times 10^{-3}} = 2000 \sin(1000 t) \, \text{m/s}^2
\]
Step 3: Maximum Speed
The maximum speed occurs when the acceleration reaches its maximum value. The maximum value of \( \sin(1000 t) \) is 1, so the maximum acceleration is:
\[
a_{\text{max}} = 2000 \, \text{m/s}^2
\]
Since the particle starts from rest, the maximum speed is attained when the particle has been accelerated for the maximum time. The velocity is the integral of acceleration:
\[
v(t) = \int a(t) \, dt = \int 2000 \sin(1000 t) \, dt
\]
The integral of \( \sin(1000 t) \) is:
\[
v(t) = -\frac{2000}{1000} \cos(1000 t) + C
\]
At \( t = 0 \), the particle is at rest, so \( C = 2 \). Thus:
\[
v(t) = 2 - 2 \cos(1000 t)
\]
The maximum speed occurs when \( \cos(1000 t) = -1 \), giving:
\[
v_{\text{max}} = 2 - 2(-1) = 4 \, \text{m/s}
\]
Step 4: Conclusion
The maximum speed attained by the particle is \( 4 \, \text{m/s} \).
Thus, the correct answer is:
\[
\boxed{(B)} \, 4 \, \text{m/s}
\]