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SCIENCE · INVESTIGATE AND EXPLAIN

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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Try

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 experiment

Predict

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.

Reset keeps your prediction, captures and practice history.

Accessible ideal trajectory

vertical displacement (mm)horizontal position (m)00.050.3750

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)
00
0.0008333330.000104167
0.001666670.000416667
0.00250.0009375
0.003333330.00166667
0.004166670.00260417
0.0050.00375
0.005833330.00510417
0.006666670.00666667
0.00750.0084375
0.008333330.0104167
0.009166670.0126042
0.010.015
0.01083330.0176042
0.01166670.0204167
0.01250.0234375
0.01333330.0266667
0.01416670.0301042
0.0150.03375
0.01583330.0376042
0.01666670.0416667
0.01750.0459375
0.01833330.0504167
0.01916670.0551042
0.020.06
0.02083330.0651042
0.02166670.0704167
0.02250.0759375
0.02333330.0816667
0.02416670.0876042
0.0250.09375
0.02583330.100104
0.02666670.106667
0.02750.113437
0.02833330.120417
0.02916670.127604
0.030.135
0.03083330.142604
0.03166670.150417
0.03250.158437
0.03333330.166667
0.03416670.175104
0.0350.18375
0.03583330.192604
0.03666670.201667
0.03750.210938
0.03833330.220417
0.03916670.230104
0.040.24
0.04083330.250104
0.04166670.260417
0.04250.270937
0.04333330.281667
0.04416670.292604
0.0450.30375
0.04583330.315104
0.04666670.326667
0.04750.338438
0.04833330.350417
0.04916670.362604
0.050.375
Calculated values under the declared model
QuantityValueUnit
Vacuum capacitance88.5419pF
Plate charge magnitude531.251pC
Uniform gap field3000N/C
Gap-to-plate-width ratio0.0141421ratio
Hypothetical vertical acceleration3000m/s²
Candidate displacement at geometric exit0.375mm
Plate reached before or at exit0flag
Accessible horizontal endpoint0.05m

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.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Derive the interior field

One ideal infinite sheet has field magnitude sigma/(2epsilon0) on either side. Equal and opposite sheets cancel outside and add inside, giving E = sigma/epsilon0. Real finite plates approximate this uniform result only sufficiently far from edges when separation is small compared with plate dimensions. Integrating across the gap gives delta V = Ed.

Separate geometry from operating state

Using Q = sigma A and C = Q/delta V gives C = epsilon0 A/d for vacuum. For a fixed linear geometry, changing Q and delta V proportionally does not change C. The displayed boundary maintains 6 V, so changing d or A changes the supplied free charge. It would be inconsistent to call that a simultaneous fixed-Q experiment.

Transfer to motion with a collision check

A passive body entering halfway between plates with horizontal speed vx and zero initial vertical velocity has ay = (q/m)E, x = vx t and y = ay t²/2. For q/m = 1 C/kg, vx = 100 m/s and a model length of 0.05 m, compute both the candidate exit time and the time to reach y = d/2. Stop at whichever occurs first. A formula evaluated beyond a plate does not establish an accessible exit state; contact dynamics are not modeled.

Reverse the signed force

Changing the sign of q/m reverses vertical deflection while the source field remains unchanged. Horizontal motion stays uniform only because the ideal field has no horizontal component and back reaction is ignored. This comparison with projectile motion depends on a uniform field; edge fields or a changing source invalidate constant acceleration.

Worked original case

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.

Investigation

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.

Changed-case practice

Try a new practice case

At maintained 6 V with a 3 mm gap, what is the gap field in N/C?

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and boundaries

Original cases checked against the official framework on 30 October 2026, with OpenStax verification on 31 October 2026 against the current official framework, Unit 10 and OpenStax capacitance. These are virtual supplied models, not instructions to charge, ground, wire or handle real equipment. Public answers are practice; retained plans and self-reports do not establish independent mastery, actual laboratory participation or AP authorization.

Apply

At maintained voltage, doubling plate area while keeping the gap fixed does what in this ideal model?

Public practice checks a specific claim. It does not establish independent mastery.

Which conductor boundary, geometric approximation or independent observation could change this conclusion?

Continue

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