Derive capacitance and bound a plate trajectory
One ideal infinite sheet has field magnitude sigma/(2epsilon0) on either side.
Learning outcome: Change area at fixed gap and predict why trajectory is unchanged although charge changes. Then change gap, compare the exit and impact conditions, and retain a rejected exit calculation. Explain the geometric limit and why the supplied q/m does not identify a real particle species.
Starting recap
Declare conductor boundaries, integrate potential or charging work, and distinguish controlled charge from voltage.
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For A = 0.02 m² and d = 2 mm, C ≈ 88.5419 pF, Q ≈ 531.251 pC and E = 3000 N/C. Candidate exit time is 0.0005 s and vertical displacement 0.375 mm, less than the 1 mm half-gap. At d = 1 mm, candidate deflection is 0.75 mm, exceeding the 0.5 mm half-gap, so the body reaches a plate first.
Open the experimentPredict
At maintained voltage, doubling plate area while keeping the gap fixed does what in this ideal model?
Experiment
Change area at fixed gap and predict why trajectory is unchanged although charge changes. Then change gap, compare the exit and impact conditions, and retain a rejected exit calculation. Explain the geometric limit and why the supplied q/m does not identify a real particle species.
Derive a plate capacitance and trajectory bound
Worked teaching model · synthetic values · Vacuum parallel plates at maintained 6 V, negligible edge field. Hypothetical passive body q/m = 1 C/kg enters midgap horizontally at 100 m/s; modeled travel length 0.05 m and initial vertical speed zero. No real particle species or physical apparatus asserted; reject exit if a plate is reached first.
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Accessible ideal trajectory
Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.
Accessible ideal trajectory: numerical coordinates
| horizontal position (m) | vertical displacement (mm) |
|---|---|
| 0 | 0 |
| 0.000833333 | 0.000104167 |
| 0.00166667 | 0.000416667 |
| 0.0025 | 0.0009375 |
| 0.00333333 | 0.00166667 |
| 0.00416667 | 0.00260417 |
| 0.005 | 0.00375 |
| 0.00583333 | 0.00510417 |
| 0.00666667 | 0.00666667 |
| 0.0075 | 0.0084375 |
| 0.00833333 | 0.0104167 |
| 0.00916667 | 0.0126042 |
| 0.01 | 0.015 |
| 0.0108333 | 0.0176042 |
| 0.0116667 | 0.0204167 |
| 0.0125 | 0.0234375 |
| 0.0133333 | 0.0266667 |
| 0.0141667 | 0.0301042 |
| 0.015 | 0.03375 |
| 0.0158333 | 0.0376042 |
| 0.0166667 | 0.0416667 |
| 0.0175 | 0.0459375 |
| 0.0183333 | 0.0504167 |
| 0.0191667 | 0.0551042 |
| 0.02 | 0.06 |
| 0.0208333 | 0.0651042 |
| 0.0216667 | 0.0704167 |
| 0.0225 | 0.0759375 |
| 0.0233333 | 0.0816667 |
| 0.0241667 | 0.0876042 |
| 0.025 | 0.09375 |
| 0.0258333 | 0.100104 |
| 0.0266667 | 0.106667 |
| 0.0275 | 0.113437 |
| 0.0283333 | 0.120417 |
| 0.0291667 | 0.127604 |
| 0.03 | 0.135 |
| 0.0308333 | 0.142604 |
| 0.0316667 | 0.150417 |
| 0.0325 | 0.158437 |
| 0.0333333 | 0.166667 |
| 0.0341667 | 0.175104 |
| 0.035 | 0.18375 |
| 0.0358333 | 0.192604 |
| 0.0366667 | 0.201667 |
| 0.0375 | 0.210938 |
| 0.0383333 | 0.220417 |
| 0.0391667 | 0.230104 |
| 0.04 | 0.24 |
| 0.0408333 | 0.250104 |
| 0.0416667 | 0.260417 |
| 0.0425 | 0.270937 |
| 0.0433333 | 0.281667 |
| 0.0441667 | 0.292604 |
| 0.045 | 0.30375 |
| 0.0458333 | 0.315104 |
| 0.0466667 | 0.326667 |
| 0.0475 | 0.338438 |
| 0.0483333 | 0.350417 |
| 0.0491667 | 0.362604 |
| 0.05 | 0.375 |
| Quantity | Value | Unit |
|---|---|---|
| Vacuum capacitance | 88.5419 | pF |
| Plate charge magnitude | 531.251 | pC |
| Uniform gap field | 3000 | N/C |
| Gap-to-plate-width ratio | 0.0141421 | ratio |
| Hypothetical vertical acceleration | 3000 | m/s² |
| Candidate displacement at geometric exit | 0.375 | mm |
| Plate reached before or at exit | 0 | flag |
| Accessible horizontal endpoint | 0.05 | m |
The ideal trajectory reaches the assigned exit before a plate.
Opposite uniform sheets cancel exterior fields and add interior fields: E = sigma/epsilon0. DeltaV = Ed then gives C = epsilon0 A/d. With passive q/m = 1 C/kg, vertical acceleration is E and horizontal speed is constant. The collision test compares displacement with half the gap; the model stops rather than inventing a rebound or continuing through a plate.
Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.
The ideal trajectory reaches the assigned exit before a plate.
Opposite uniform sheets cancel exterior fields and add interior fields: E = sigma/epsilon0. DeltaV = Ed then gives C = epsilon0 A/d. With passive q/m = 1 C/kg, vertical acceleration is E and horizontal speed is constant. The collision test compares displacement with half the gap; the model stops rather than inventing a rebound or continuing through a plate.
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