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SCIENCE · INVESTIGATE AND EXPLAIN

End an electric trajectory at its first boundary event

A uniform electric field gives a passive charge constant force qE. In this fictional model the bead has q=1 nC and mass 10⁻⁸ kg, enters halfway between plates with horizontal velocity and zero vertical velocity, and experiences no gravity. The vertical acceleration is a_y=qE/m=0.1E in SI units. Horizontal velocity stays constant because the ideal field is vertical.

Learning outcome: Compare E=−100,0,+100 N/C at fixed horizontal speed. Then increase speed to 3 m/s and classify the first event again. Retain event times, endpoint and work/energy residual rather than extrapolating all paths to the far edge.

Starting recap

Use signed charge, common coordinate axes, SI units, vector components and algebraic ratios. Separate a source field from probe force and a fitted parameter from independently tested evidence.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

For E=100 N/C and vx=2 m/s, a_y=10 m/s². Contact occurs at √.02≈.141421 s before exit at .2 s. At contact x≈.282843 m, y=.1 m and vy≈1.414214 m/s. Work qEΔy=10⁻⁸ J matches the gained vertical kinetic energy.

Open the experiment

Predict

If plate contact occurs before horizontal exit, the valid endpoint is:

Experiment

Compare E=−100,0,+100 N/C at fixed horizontal speed. Then increase speed to 3 m/s and classify the first event again. Retain event times, endpoint and work/energy residual rather than extrapolating all paths to the far edge.

Stop the trajectory at the first plate or exit event

Worked teaching model · synthetic values · Fictional bead q=1 nC, mass=1e−8 kg enters at y=0 with vy=0. Plates at y=±.1 m, length .4 m, uniform field inside and no gravity. A first contact or horizontal exit ends the model; no post-contact or fringe prediction. Virtual only.

Reset keeps your prediction, captures and practice history.

Trajectory only up to first event

y (m)x (m)00.2828430.10

Spatial axes use the same pixels per physical unit, preserving geometry. 1: bounded bead trajectory (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Trajectory only up to first event

bounded bead trajectory

x (m)y (m)
00
0.007071070.0000625
0.01414210.00025
0.02121320.0005625
0.02828430.001
0.03535530.0015625
0.04242640.00225
0.04949750.0030625
0.05656850.004
0.06363960.0050625
0.07071070.00625
0.07778170.0075625
0.08485280.009
0.09192390.0105625
0.09899490.01225
0.1060660.0140625
0.1131370.016
0.1202080.0180625
0.1272790.02025
0.134350.0225625
0.1414210.025
0.1484920.0275625
0.1555630.03025
0.1626350.0330625
0.1697060.036
0.1767770.0390625
0.1838480.04225
0.1909190.0455625
0.197990.049
0.2050610.0525625
0.2121320.05625
0.2192030.0600625
0.2262740.064
0.2333450.0680625
0.2404160.07225
0.2474870.0765625
0.2545580.081
0.261630.0855625
0.2687010.09025
0.2757720.0950625
0.2828430.1

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Vertical acceleration10m/s²
First event time0.141421s
First event x0.282843m
First event y0.1m
First event is contact (includes tie)1flag
Horizontal velocity at event2m/s
Vertical velocity at event1.41421m/s
Kinetic energy gain1e-8J
Electric work to event1e-8J
Potential difference to event-10V

The first plate contact ends the model before any later exit prediction.

Use a_y=qE/m, y=a_y t²/2 and x=v_x t. Compare event times before evaluating endpoints; the zero-field case has no plate-contact event. Work qE Delta y equals the kinetic-energy increase, with Delta V=−E Delta y. Contact is not a continuation through a plate; a simultaneous exit/contact is labeled explicitly by the contact flag.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The first plate contact ends the model before any later exit prediction.

Use a_y=qE/m, y=a_y t²/2 and x=v_x t. Compare event times before evaluating endpoints; the zero-field case has no plate-contact event. Work qE Delta y equals the kinetic-energy increase, with Delta V=−E Delta y. Contact is not a continuation through a plate; a simultaneous exit/contact is labeled explicitly by the contact flag.

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

Understand

Define the source and measurement boundary

A uniform electric field gives a passive charge constant force qE. In this fictional model the bead has q=1 nC and mass 10⁻⁸ kg, enters halfway between plates with horizontal velocity and zero vertical velocity, and experiences no gravity. The vertical acceleration is a_y=qE/m=0.1E in SI units. Horizontal velocity stays constant because the ideal field is vertical.

Derive the quantitative comparison

The plate region has length .4 m and surfaces at y=±.1 m. Horizontal exit time is .4/vx. When E is nonzero, contact time is √(.2/|a_y|), obtained from |y|=|a_y|t²/2. Evaluate the trajectory only up to the earlier event. A calculation that continues past the first plate contact is not an alternative valid prediction: it has left its own physical domain. At zero field there is no plate-contact event in this model.

Test an alternative before claiming identification

The endpoint energy check is independent of the time-root procedure: electric work qEΔy equals the gain in kinetic energy, and ΔV=−EΔy. Reversing E reverses vertical displacement and velocity but leaves work positive for this initially un-deflected positive bead, because the product EΔy stays positive. A negative charge would instead reverse the acceleration at fixed field. Fringe fields, contact forces and motion after exit are absent; no actual charged particle apparatus or electrical experiment is requested.

Worked original case

For E=100 N/C and vx=2 m/s, a_y=10 m/s². Contact occurs at √.02≈.141421 s before exit at .2 s. At contact x≈.282843 m, y=.1 m and vy≈1.414214 m/s. Work qEΔy=10⁻⁸ J matches the gained vertical kinetic energy.

Investigation

Compare E=−100,0,+100 N/C at fixed horizontal speed. Then increase speed to 3 m/s and classify the first event again. Retain event times, endpoint and work/energy residual rather than extrapolating all paths to the far edge.

Changed-case practice

Try the changed case before feedback

For E=0 and horizontal speed 2 m/s, find the exit time in seconds.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and evidence boundary

Original teaching, numerical cases and diagrams checked 30 October 2026 against the relevant primary textbook explanation. No textbook exercise, image or passage is reproduced. These are bounded synthetic records, not observed apparatus. No charging, wiring, capacitor handling, medical application or safety certification is requested. Preserve first predictions, contrasting settings and critiques in the course evidence workspace; public feedback is not independent mastery or performed laboratory time.

Apply

If plate contact occurs before horizontal exit, the valid endpoint is:

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

Which independent observation could reject this model rather than merely refit its parameters?

Continue

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