📺Charged Particle Deflection - CRT, Electric Field Deflection, AC Plates

How to use Charged Particle Acceleration & Deflection

1. 1Pick a particle (electron, proton, deuteron, alpha) and set U₁, U₂, plate length L, gap d and screen distance D
2. 2Press Play to watch it accelerate, bend and hit the screen; read v₀, offset y, tanθ, screen offset Y and the critical voltage
3. 3Switch to the CRT or AC tab and change the sweep, the signal, the period and the entry time to see the screen picture and exit offsets
4. 4Try the random problems on the Practice tab and check the worked solution

FAQ

Why does the offset y = U₂L²/(4U₁d) not depend on charge or mass?

Acceleration gives qU₁ = ½mv₀², so mv₀² = 2qU₁. Between the plates y = ½(qU₂/md)(L/v₀)² = qU₂L²/(2mdv₀²). Substituting mv₀² cancels both q and m, leaving y = U₂L²/(4U₁d). Ions of the same sign accelerated from rest by the same voltage and sent through the same plates therefore follow identical paths; they just take different times.

Why does the exit velocity line pass through the centre of the plates?

At the exit the offset is y = ½at², the vertical velocity is v_y = at and the horizontal distance is L = v₀t. Extending the exit velocity backwards covers a horizontal distance y/tanθ = (½at²)/(at/v₀) = v₀t/2 = L/2. So the particle seems to come straight from the plate centre, and the screen offset is simply Y = (L/2 + D)tanθ.

How do I know whether the particle hits a plate?

For entry along the centre line it hits when the computed offset exceeds d/2. Setting y = d/2 gives the critical voltage U₂ = 2U₁d²/L². With a larger U₂ the particle lands at x = d√(2U₁/U₂) from the entrance. The tool lists the critical voltage and marks the landing point on the canvas.

Why does an oscilloscope put a sawtooth on the X plates?

A sawtooth rises linearly, moving the spot across the screen at a steady speed, which turns the horizontal axis into a time axis; then it snaps back and the next sweep starts. The signal on the Y plates moves the spot up and down at the same time, so the screen draws the signal against time. When the sweep period is a whole number of signal periods every sweep lands on the same curve and the picture is stable.

When do particles leave alternating plates parallel to them?

When the transit time is a whole number of voltage periods. With a square wave each period pushes the particle one way for half a period and brakes it for the other half, so the vertical velocity returns to zero and every particle leaves parallel to the plates. The one entering at t = 0 is displaced most, y = aT²/4 per period; drag the entry-time slider to see the offset change.