describe the effect of a uniform electric field on the motion of charged particles

Uniform Electric Fields

Objective

Describe the effect of a uniform electric field on the motion of charged particles.

Key Concepts

• A uniform electric field has the same magnitude and direction at every point in the region.
• Field lines are parallel, equally spaced, and point from positive to negative potential.
• The electric field strength \$E\$ is related to the potential difference \$V\$ and the separation \$d\$ by \$E = \dfrac{V}{d}\$.
• The force on a charge \$q\$ in an electric field is \$ \mathbf{F} = q\mathbf{E}\$.
• Resulting acceleration is \$ \mathbf{a} = \dfrac{q\mathbf{E}}{m}\$, where \$m\$ is the particle’s mass.

Mathematical Description

For a particle of charge \$q\$ and mass \$m\$ placed in a uniform electric field \$\mathbf{E}\$:

\$\mathbf{F}=q\mathbf{E}\$

\$\mathbf{a}=\frac{\mathbf{F}}{m}=\frac{q\mathbf{E}}{m}\$

The direction of \$\mathbf{a}\$ is the same as \$\mathbf{E}\$ for a positive charge and opposite for a negative charge.

Motion of Charged Particles

1. Field Parallel to Initial \cdot elocity

• If the particle’s initial velocity \$\mathbf{v}_0\$ is parallel (or anti‑parallel) to \$\mathbf{E}\$, the motion is one‑dimensional.
• Equations of motion:

  \$v = v_0 + at\$

  \$x = v_0 t + \frac{1}{2} a t^2\$

• A positive charge accelerates in the direction of the field; a negative charge decelerates (or accelerates opposite to the field).

2. Field Perpendicular to Initial \cdot elocity

• When \$\mathbf{v}_0\$ is perpendicular to \$\mathbf{E}\$, the particle experiences constant acceleration in the direction of the field while moving uniformly in the perpendicular direction.
• Resulting trajectory is a parabola, analogous to projectile motion under gravity:

  \$x = v_{0x} t\$

  \$y = \frac{1}{2} a t^2\$

  where \$a = \dfrac{qE}{m}\$.

• The path lies in a plane defined by \$\mathbf{v}_0\$ and \$\mathbf{E}\$.

3. General Case – Arbitrary Initial Direction

1. Resolve the initial velocity into components parallel (\$v{0\parallel}\$) and perpendicular (\$v{0\perp}\$) to \$\mathbf{E}\$.
2. Apply the one‑dimensional equations to each component:

   • Parallel component: \$v{\parallel}=v{0\parallel}+at\$, \$x{\parallel}=v{0\parallel}t+\frac12 a t^2\$.
   • Perpendicular component: \$v{\perp}=v{0\perp}\$ (constant), \$x{\perp}=v{0\perp}t\$.

3. The overall trajectory is a combination of uniform motion in the perpendicular direction and uniformly accelerated motion in the parallel direction, producing a curved path.

Energy Considerations

The work done by the electric field changes the particle’s kinetic energy:

\$W = qE d = \Delta K = \frac{1}{2} m vf^2 - \frac{1}{2} m vi^2\$

Equivalently, the change in electric potential energy is

\$\Delta U = q \Delta V = -W\$

Common Misconceptions

• “Electric fields always accelerate particles.” – Only the component of the field parallel to the charge’s motion changes the speed; perpendicular components alter direction.
• “The field does work on a neutral particle.” – A neutral particle experiences no net electric force, so the field does no work on it.
• “All charged particles follow the same path in a uniform field.” – The trajectory depends on charge sign, magnitude, mass, and initial velocity.

Summary Table