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SCIENCE · INVESTIGATE AND EXPLAIN

Distinguish a field null from a zero potential

For two positive point charges on a line, fields point away from each source. Between them the contributions oppose; outside them they point in the same direction. Therefore the finite cancellation point lies between the sources, not beyond an endpoint. Source positions themselves are singular and cannot be substituted as candidate nulls.

Learning outcome: Hold separation at 2 m and compare right charges 1,5,9 nC. Predict the direction of null motion before checking. Draw the separate equal-and-opposite midpoint case and use vector arrows and scalar signs to explain its contrasting cancellations.

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 a=9 and d=2 m, the field null is x=2/(1+3)=.5 m. Its distances from the sources are .5 and 1.5 m. The source-field magnitudes match because 1/.5²=9/1.5²=4; the potential contributions are both positive.

Open the experiment

Predict

At the null between two positive sources, potential relative to infinity is:

Experiment

Hold separation at 2 m and compare right charges 1,5,9 nC. Predict the direction of null motion before checking. Draw the separate equal-and-opposite midpoint case and use vector arrows and scalar signs to explain its contrasting cancellations.

Find a field null without confusing potential cancellation

Worked teaching model · synthetic values · Left source +1 nC at x=0 and right positive source at x=d. One field null lies strictly between them. Compare the midpoint potential of a separately stipulated equal-and-opposite pair; source locations themselves are excluded.

Reset keeps your prediction, captures and practice history.

Same-sign potential along interior segment

potential (V)x (m)01.8229.6820

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: sum of source potentials (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Same-sign potential along interior segment

sum of source potentials

x (m)potential (V)
0.269.9032
0.2462.981
0.2858.225
0.3254.8348
0.3652.3665
0.450.555
0.4449.2325
0.4848.2884
0.5247.6471
0.5647.256
0.647.0777
0.6447.0855
0.6847.2607
0.7247.5903
0.7648.0659
0.848.6826
0.8449.4389
0.8850.3361
0.9251.3781
0.9652.5714
153.9253
1.0455.452
1.0857.1672
1.1259.0902
1.1661.2452
1.263.6618
1.2466.3767
1.2869.4351
1.3272.8937
1.3676.8237
1.481.3159
1.4486.4874
1.4892.4914
1.5299.5332
1.56107.893
1.6117.962
1.64130.307
1.68145.78
1.72165.717
1.76192.347
1.8229.682

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
Field-null coordinate0.618034m
Distance from right source1.38197m
Potential at field null47.0594V
Net field at declared null0N/C
Equal-opposite comparison midpoint potential0V
Equal-opposite comparison midpoint field17.9751N/C

Zero electric field need not mean zero potential, and zero potential need not mean zero field.

Between positive sources fields oppose: 1/x²=a/(d−x)² gives x=d/(1+sqrt(a)). Both potentials are positive relative to infinity, so their sum cannot vanish. A separate +1/−1 nC pair has zero midpoint potential but reinforcing rightward fields. Cancellation claims must identify vector versus scalar quantity and the source configuration.

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

Zero electric field need not mean zero potential, and zero potential need not mean zero field.

Between positive sources fields oppose: 1/x²=a/(d−x)² gives x=d/(1+sqrt(a)). Both potentials are positive relative to infinity, so their sum cannot vanish. A separate +1/−1 nC pair has zero midpoint potential but reinforcing rightward fields. Cancellation claims must identify vector versus scalar quantity and the source configuration.

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

Understand

Define the source and measurement boundary

For two positive point charges on a line, fields point away from each source. Between them the contributions oppose; outside them they point in the same direction. Therefore the finite cancellation point lies between the sources, not beyond an endpoint. Source positions themselves are singular and cannot be substituted as candidate nulls.

Derive the quantitative comparison

Let the left charge be +1 nC at x=0 and the right be +a nC at x=d. Equating magnitudes gives 1/x²=a/(d−x)². Both distances are positive inside the interval, so d−x=√a x and x=d/(1+√a). The stronger source lies farther from the null. The field is zero there, but V=k×10⁻⁹[1/x+a/(d−x)] remains positive relative to infinity.

Test an alternative before claiming identification

Compare a separate +1/−1 nC pair at the same positions. At its midpoint the signed scalar potentials cancel, yet both field contributions point from the positive source toward the negative one. Their sum is 8k×10⁻⁹/d². Zero field, zero potential, zero total charge and zero potential energy are different statements. The potential reference matters for V=0; adding a constant changes that statement but not any force or potential difference. A field null also does not by itself prove stable equilibrium for arbitrary three-dimensional displacements.

Worked original case

For a=9 and d=2 m, the field null is x=2/(1+3)=.5 m. Its distances from the sources are .5 and 1.5 m. The source-field magnitudes match because 1/.5²=9/1.5²=4; the potential contributions are both positive.

Investigation

Hold separation at 2 m and compare right charges 1,5,9 nC. Predict the direction of null motion before checking. Draw the separate equal-and-opposite midpoint case and use vector arrows and scalar signs to explain its contrasting cancellations.

Changed-case practice

Try the changed case before feedback

Left charge +1 nC and right +4 nC are separated by 3 m. Find the field-null distance from the left source in metres.

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

At the null between two positive sources, potential relative to infinity 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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