Motion of Charged Particles in Electric Fields

Expert reviewed 22 November 2024 6 minute read


Introduction

When charged particles enter electric fields, they experience forces that can dramatically alter their motion. Understanding these interactions is crucial for many applications in modern physics and technology, from particle accelerators to television displays.

Electric Fields Between Parallel Plates

A uniform electric field is created when two parallel conducting plates are connected to a voltage source. The strength of this electric field (E) is given by:

E=VdE = \frac{V}{d}

Where:

  • E is the electric field strength (measured in V/m or N/C)
  • V is the potential difference between the plates (in volts)
  • d is the separation distance between plates (in meters)
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Forces on Charged Particles

When a charged particle enters an electric field, it experiences an electric force given by:

F=qEF = qE

Where:

  • F is the force (in newtons)
  • q is the particle's charge (in coulombs)
  • E is the electric field strength

This force causes acceleration according to Newton's Second Law:

F=maF = ma

Therefore, the acceleration of the charged particle is:

a=qEma = \frac{qE}{m}

Trajectories in Electric Fields

When a charged particle enters an electric field with an initial velocity, its motion can be analyzed by breaking it down into two components:

  • Motion parallel to the electric field: Experiences constant acceleration
  • Motion perpendicular to the electric field: Maintains constant velocity

This results in a parabolic trajectory, similar to projectile motion in gravitational fields.

Comparison with Gravitational Fields

CharacteristicElectric FieldsGravitational Fields
Force DirectionDepends on charge polarityAlways attractive
Force MagnitudeF=qEF = qEF=mgF = mg
Trajectory ShapeParabolicParabolic
AccelerationUsually larger9.81 m/s² on Earth
Affected ObjectsOnly charged particlesAll matter

Negligible Gravitational Effects

For charged particles, gravitational forces are typically negligible compared to electric forces. For example, an electron near Earth's surface experiences a gravitational force of:

Fg=GMmr2F_g = \frac{GMm}{r^2}

Fg=(6.67×1011)(6.0×1024)(9.109×1031)(6.371×106)2=9.0×1030 NF_g = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(9.109 \times 10^{-31})}{(6.371 \times 10^6)^2}\\ = 9.0 \times 10^{-30} \text{ N}

This is vastly smaller than typical electric forces on the same particle.

Practice Question 1

Consider a proton entering an electric field with:
  • Potential difference: 150 V
  • Plate separation: 2.0 m

Find the electric field strength and force acting on the proton.

The electric field strength is:

E=Vd=150 V2.0 m=75 V/mE = \frac{V}{d} = \frac{150 \text{ V}}{2.0 \text{ m}} = 75 \text{ V/m}

The force on the proton is:

F=qE=(1.60×1019 C)(75 V/m)=1.20×1017 NF = qE = (1.60 \times 10^{-19} \text{ C})(75 \text{ V/m}) \\= 1.20 \times 10^{-17} \text{ N}

Return to Module 6: Electromagnetism