= Solution
A charge $q$ moving nonrelativistically in prescribed fields obeys the <Lorentz force law>
$$
\boxed{m\frac{d\mathbf v}{dt}=q(\mathbf E+\mathbf v\times\mathbf B).}
$$
The magnetic part is perpendicular to <velocity>, so $d(mv^2/2)/dt=q\mathbf E\cdot\mathbf v$: the <electric field> can change <kinetic energy>, while the <magnetic field> alone changes direction without doing <work>.
In a uniform <electric field> with no <magnetic field>, <acceleration> is constant: $\mathbf v=\mathbf v_0+(q\mathbf E/m)t$ and $\mathbf r=\mathbf r_0+\mathbf v_0t+q\mathbf E t^2/(2m)$. The trajectory is a parabola when there is initial motion perpendicular to the field, and a line when motion is purely along it. Reversing the charge reverses the electric <acceleration>.
In a uniform <magnetic field> with no <electric field>, resolve <velocity> into parallel and perpendicular components. The parallel component is constant; the perpendicular component rotates at the <cyclotron frequency>
$$
\boxed{\omega_c=\frac{|q|B}{m},\qquad r_L=\frac{mv_\perp}{|q|B}.}
$$
The resulting path is a helix, or a circle if $v_\parallel=0$, with pitch $2\pi v_\parallel/\omega_c$. Charge sign determines its sense of rotation. These relations underlie magnetic bending and <mass>/charge analysis. At relativistic energies, use <momentum> $\mathbf p=\gamma m\mathbf v$; at fixed <energy> the <cyclotron frequency> becomes $|q|B/(\gamma m)$.
For uniform crossed fields $\mathbf E\perp\mathbf B$, the <velocity> $\mathbf v_D=\mathbf E\times\mathbf B/B^2$ satisfies $\mathbf E+\mathbf v_D\times\mathbf B=0$. The orbit is a gyromotion about a <guiding center> drifting with
$$
\boxed{\mathbf v_D=\frac{\mathbf E\times\mathbf B}{B^2},}
$$
independent of <mass> and charge sign. A <velocity selector> transmits particles with the appropriate direction and <speed> $E/B$ without deflection. If the <electric field> has a component along the <magnetic field>, that component accelerates the parallel motion instead. Nonuniform fields can add other drifts; the simple helix and uniform crossed-field expressions require the stated uniform-field assumptions.
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