The influence of the various parameters on the movement path of a charged particle as an electron can be quite "tricky".
On should reflect wll on what one sees. Some hints:
No magnetic and electric field: the particle will simply follow its initial velocity vector. Its acceleration is zero.
Electric field only: The electric field vector determines acceleration
AccelerationEx ≈ Ex
With constant electric field a particle has constant acceleration in direction of the electric field vector. Its velocity will increase linearly.
Magnetic field only: The acceleration vector is determined by the vector product of velocity vector and magnetic field vector. It is perpendicular to the plane of both vectors. Its value is proportional to the sine of the angle between the velocity vector and the field vector.
AccelerationBx = (v x B )x ≈ (vyBz - vzBy)
A charge at rest (vy = 0; vz = 0) will not be influenced by the magnetic field, as is one which is moving in the direction of the magnetic field (sin... =0)
When the velocity vector is oblique to the field vector, the particle path will be a screw of constant pitch around an axis parallel to the magnetic field vector. When the velocity vector is perpendicular to the magnetic field vector, the path will be a circle in a plane perpendicular to the field vector.
Both electric and magnetic field: The accelerations add. In general one will get a screw path with increasing or decreasing pitch:
Acceleration(B&E)x = ≈ Ex+(vyBz - vzBy)
Acceleration(B&E)y = ≈ Ey+(vzBx - vxBz)
Acceleration(B&E)z = ≈ Ez+(vzBx - vxBz)
The following observation is quite surprising: with an electric field perpendicular to the magnetic field the particle seems to circle around the magnetic field vector, but is slowly wandering perpendicular to the electric field vector. One should recognize that in this case the particle is accelerated by the electric field during one half revolution and decelerated in the second one. It is not really moving in a circle!