A particle of charge q and mass m released from origin with velocity `vec(v) = v_(0) hat(i)` into a region of uniform electric and magnetic fields parallel to y-axis. i.e., `vec(E) = E_(0) hat(j) and vec(B) = B_(0) hat(j)`. Find out the position of the particle as a functions of time
Strategy : Here `vec(E) || vec(B)`
The electric field accelerates the particle in y-direction i.e., component of velocity goes on increasing with acceleration
`a_(y) = (F_(y))/(m) = (F_(e))/(m) = (qE_(0))/(m)`
The magnetic field rotates the particle in a circle in x-z plane (perpendicular to magnetic field) The resultant path of the particle is a helix with increasing pitch.
Velocity of the particle at time t would be
`vec(v) (t) = v_(x) hat(i) + v_(y) hat(j) + v_(z) hat(k)`

Explanation

To solve the problem of a charged particle moving in uniform electric and magnetic fields, we will break down the solution into manageable steps.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle:**
The particle has a charge \( q \) and mass \( m \). It is subjected to an electric field \( \vec{E} = E_0 \hat{j} \) and a magnetic field \( \vec{B} = B_0 \hat{j} \). The electric force acting on the particle is given by:
\[
\vec{F}_e = q \vec{E} = q E_0 \hat{j}
...