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:

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

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

Initial \cdot elocity Direction	Resulting Motion	Key Equations
Parallel to $\mathbf{E}$	Linear acceleration (or deceleration) along field lines	$v = v_0 + \dfrac{qE}{m}t$, $x = v_0 t + \dfrac{1}{2}\dfrac{qE}{m}t^2$
Perpendicular to $\mathbf{E}$	Parabolic trajectory (uniform motion sideways, accelerated motion along field)	$x = v_{0x}t$, $y = \dfrac{1}{2}\dfrac{qE}{m}t^2$
Arbitrary angle	Combination of uniform and accelerated components → curved path	Resolve into components; apply component equations separately