Skip to main content
← Electric Fields

Subject 11 · Free game, no account

Field Runner

Drag the plates, dial the volts, land the electron in the slot.

E = V/da = qE/my = ½at²

Field Runner

Score0best 0

++++E = 1.7 kV/m e⁻SLOT+32 mm17.9 mm too lowDrag the plates apartchamber 320 × 200 mm · grid 40 mm · dashed line is the undeflected axisv₀ = 2.0 × 10⁷ m/s

Run 1/4First Deflection

E = V/d1.67kV/m
a = qE/m2.93 × 10¹⁴m/s²

Land the beam in the slot 32 mm above the axis.

Simple definition
A uniform electric field has the same magnitude and direction everywhere in a region; ideal parallel plates produce one between them.
Example
400 V across a 20 mm gap makes a 20 kV/m field, and an electron in it accelerates at 3.5 × 10¹⁵ m/s² toward the positive plate.
How the physics works

Between wide, closely spaced plates the field is very nearly uniform and equal to the potential difference divided by the separation, E = V/d. The field is defined by the force per unit positive charge, so a particle of charge q feels F⃗ = qE⃗ and accelerates at a⃗ = qE⃗/m. E⃗ points from the positive plate to the negative one, and because the electron is negative it is pushed the other way — toward the positive plate.

The field has no component along the beam, so the horizontal speed never changes and the time inside the plates is t = L/v₀. The transverse motion is uniform acceleration from rest, y = ½at², which is the parabola you can see bending inside the stack. After the plates the particle travels in a straight line, so the drift multiplies the exit angle by the remaining distance.

Halving d doubles E at fixed voltage and so doubles the deflection — but it also halves the corridor to d/2. The share of the total deflection that has to happen between the plates is fixed by geometry alone, so on every run there is a smallest gap that can possibly work. That is the tension the second run is built on.

The bead on the last run is a point charge, so its field is radial and falls as the inverse square of distance, E = kQ/r². Fields superpose as vectors, so its contribution simply adds to the plate term — but it does not scale with V, which is why the last run needs a genuinely different setting rather than a rescaled one.

  • e = 1.602 × 10⁻¹⁹ C elementary charge
  • m = 9.109 × 10⁻³¹ kg electron rest mass
  • k = 8.99 × 10⁹ N·m²/C² Coulomb constant, k = 1/(4πε₀)
  • v₀ = 2.0 × 10⁷ m/s fixed gun speed, 0.067 c
  • Semi-implicit Euler, 4000 steps across a 320 mm chamber
  • Ignored plate-edge fringing, gravity, radiation and relativity

Game 11 · Electric Fields learning guide

Turn the playthrough into a physics lesson.

Learning objectiveSet plate voltage and separation to control an electron’s acceleration and land it in a target slot.

01

What you will learn

  • A uniform field between plates has magnitude |ΔV|/d.
  • A negative charge accelerates opposite to the electric field.
  • Particle motion combines constant horizontal velocity with vertical acceleration.

02

How to play

  1. Read the launch conditions and target position.
  2. Adjust the plate voltage and gap while watching the predicted trajectory.
  3. Fire the electron and refine the field until it reaches the slot.

03

Quick classroom check

What happens to field strength when the same potential difference is applied across twice the plate separation?

Suitable forUpper-secondary physics · uniform electric fields

Continue this topic

Move from play to explanation and exam-style practice.

Teachers can share the page link with a class. The game is free and does not require an account.