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Physics HL · Chapter 19: Motion in Electric and Magnetic Fields

19.1 Motion in Uniform Electric Fields

Model charged-particle motion between parallel plates using constant-acceleration mechanics and energy-potential links.

Estimated time: 42 minutes

Force and Acceleration Along a Field Line

In a uniform electric field, the electric force on a particle is constant in magnitude and direction as long as the particle stays inside the uniform region. That means acceleration is constant too. For a positive charge, acceleration is along E. For a negative charge, acceleration is opposite E. This is the exact same kinematic structure as constant-acceleration motion in Chapter 1.

F_E = qE,qquad a = rac{qE}{m}

Sign of q controls acceleration direction; magnitude scales with the charge-to-mass ratio q/m.

If the particle starts from rest, or if initial velocity is exactly parallel to E, the path stays straight. Speed changes because force has a component along velocity. This is the fastest case to solve: one-dimensional constant acceleration with clear sign convention.

Perpendicular Entry: Why the Path Becomes Parabolic

When a particle enters the field with horizontal velocity while E is vertical, there is no horizontal force in the ideal model. Horizontal speed stays constant, while vertical velocity changes uniformly. That is mathematically identical to horizontal projectile launch with gravity replaced by qE/m, so the trajectory is parabolic.

ight)t^2,qquad v_y = u_y + left( rac{qE}{m} ight)t$$

Use component form directly; do not force a single-vector equation when only one component accelerates.

This component strategy is also the easiest way to decide whether a particle will strike one plate before exiting. Compute the transit time from plate length and horizontal speed, then evaluate vertical displacement at that time. If displacement magnitude exceeds half-gap, the particle collides with a plate before leaving the field region.

Energy View: Work by the Electric Field

Unlike magnetic force, electric force can do work because it can have a component along displacement. In uniform fields between plates, energy changes are often faster to compute from potential difference than from acceleration-time methods. This provides an independent check on kinematics-based answers.

DeltaEk=qDeltaV=qEDeltasparallelEDelta E_k = qDelta V = qEDelta s_{parallel E}

A positive Delta Ek means the field transferred energy into particle kinetic energy.

Simulation: Electric-Field Trajectory Studio

Launch electrons, protons, or alpha particles between plates and test when trajectories stay clear, strike plates, or exit with strong angle change.

plateplateinitial velocity (green), final velocity (orange), trajectory (yellow)

Vertical acceleration

2.46e+14 m/s^2

Time in field

1.31e-8 s

Vertical displacement

2.40 cm

Exit angle

27.94 deg

Delta kinetic energy

5.38e-18 J

Plate status

Hits upper plate

Final speed

7.35e+6 m/s

Work-energy check

Field does work

Test Yourself

A proton is accelerated from rest through 120 V. Enter its speed after acceleration.

Hint: Use qV = (1/2)mv^2 with q = 1.60 x 10^-19 C and m = 1.67 x 10^-27 kg.