1. Given:

   • Electric field $\vec{E}$ is steady and uniform.
   • Magnetic field $\vec{B}$ is steady and uniform.
   • $\vec{E}$ and $\vec{B}$ are parallel.
   • A charged particle is released from rest.

2. Forces acting on the charged particle: The Lorentz force is

   $$\vec{F} = q\left(\vec{E} + \vec{v} \times \vec{B}\right).$$

3. Initial condition: Since the particle is released from rest,

   $$\vec{v}=0.$$

   Therefore initially,

   $$\vec{v} \times \vec{B} = 0.$$

   So the only force initially is electric force:

   $$\vec{F} = q\vec{E}.$$

4. Direction of motion: The particle starts accelerating along the direction of $\vec{E}$ (or opposite to it if $q<0$). Since $\vec{B}$ is parallel to $\vec{E}$, the velocity $\vec{v}$ remains along the same line as $\vec{B}$.

5. Magnetic force during motion: At any later time, $\vec{v}$ is parallel (or antiparallel) to $\vec{B}$. Hence,

   $$\vec{v} \times \vec{B} = 0.$$

   So magnetic force is always zero.

6. Nature of path: Since only the electric force acts, and it is along a fixed direction, the particle moves along a straight line.

7. Option check:

   • A: helix $\rightarrow$ requires a velocity component perpendicular to $\vec{B}$, not possible here.
   • B: straight line $\rightarrow$ correct.
   • C: ellipse $\rightarrow$ not possible under constant force in one direction.
   • D: circle $\rightarrow$ requires magnetic force perpendicular to velocity, not possible here.

Therefore, the correct answer is