SCIENCE · INVESTIGATE AND EXPLAIN

Distinguish equal potential from equal field

Use scalar superposition and vector projection as different measurements of a source.

Learning outcome: Compare center and axial records at two radii. Predict which quantities scale as inverse radius and inverse radius squared. Design a two-observation test for ring versus arc without claiming a single scalar value identifies the source.

Starting recap

Keep source normalization, potential reference, signed energy and evidence independence explicit.

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Try

At R = 1 m, z = 1 m and Q = 2 nC, V ≈ 12.7103 V and Ez ≈ 6.35516 N/C. At the center V ≈ 17.9751 V while the ring field is zero. Replacing the ring by an equal-charge semicircle leaves that center potential unchanged but creates a nonzero field.

Open the experiment

Predict

Equal charge on a ring and a semicircle of equal radius gives the same center potential. What follows about center field?

Your prediction and reason

Experiment

Compare center and axial records at two radii. Predict which quantities scale as inverse radius and inverse radius squared. Design a two-observation test for ring versus arc without claiming a single scalar value identifies the source.

Compare scalar potential with a vector field

Worked teaching model · synthetic values · Thin uniform ring Q = 2 nC. V(infinity) = 0; axis only. Separate equal-Q arc-center comparisons use the same radius, not the selected axial coordinate. Identical center potential does not imply identical center field.

Ring radius: 1 m Axial observation coordinate: 1 m

Reset keeps your prediction, captures and practice history.

Ring-axis potential

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.

Ring-axis potential: numerical coordinates

z (m)	V (V)
-3	5.68423
-2.9	5.85972
-2.8	6.04568
-2.7	6.24301
-2.6	6.45269
-2.5	6.67579
-2.4	6.9135
-2.3	7.16714
-2.2	7.43815
-2.1	7.7281
-2	8.03871
-1.9	8.37184
-1.8	8.72948
-1.7	9.11374
-1.6	9.52679
-1.5	9.97079
-1.4	10.4478
-1.3	10.9596
-1.2	11.5074
-1.1	12.0914
-1	12.7103
-0.9	13.3608
-0.8	14.0362
-0.7	14.7258
-0.6	15.4135
-0.5	16.0774
-0.4	16.6895
-0.3	17.217
-0.2	17.626
-0.1	17.8859
0	17.9751
0.1	17.8859
0.2	17.626
0.3	17.217
0.4	16.6895
0.5	16.0774
0.6	15.4135
0.7	14.7258
0.8	14.0362
0.9	13.3608
1	12.7103
1.1	12.0914
1.2	11.5074
1.3	10.9596
1.4	10.4478
1.5	9.97079
1.6	9.52679
1.7	9.11374
1.8	8.72948
1.9	8.37184
2	8.03871
2.1	7.7281
2.2	7.43815
2.3	7.16714
2.4	6.9135
2.5	6.67579
2.6	6.45269
2.7	6.24301
2.8	6.04568
2.9	5.85972
3	5.68423

Calculated values under the declared model
Quantity	Value	Unit
Ring axial potential	12.7103	V
Ring axial field	6.35516	N/C
Ring and equal-Q arc center potential	17.9751	V
Uniform ring center field magnitude	0	N/C
Semicircle center field magnitude at same Q and radius	11.4433	N/C
Quarter-arc center field magnitude at same Q and radius	16.1833	N/C

Equal potential at one location can coexist with very different field vectors there.

Every ring element is at the same axial distance, so scalar integration gives kQ/sqrt(R² + z²). Every arc element is distance R from its center, so center V = kQ/R regardless of span at fixed Q. Field uses vector projection and therefore retains angular-distribution information that one potential value loses.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Equal potential at one location can coexist with very different field vectors there.

Observation to attach to a capture

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

Understand

Integrate the scalar quantity

Every element of a ring of radius R is the same distance sqrt(R² + z²) from an axial observer. Therefore V(z) = kQ/sqrt(R² + z²). Differentiation gives Ez = kQz/(R² + z²)^(3/2). The potential is even in z and the axial field odd. A stationary potential at the center is compatible with a nonzero potential value.

Construct a degenerate observation

Every charge element on any arc of the same circle is distance R from the center, so Vcenter = kQ/R at fixed Q. A full uniform ring, a semicircle and a quarter-circle can share that center potential while having different fields. The ring field cancels; the semicircle magnitude is 2kQ/(pi R²), and the quarter-circle magnitude is 2sqrt(2)kQ/(pi R²). Those comparisons hold total charge fixed, not linear density.

Choose a discriminating observation

A single center potential cannot identify angular distribution. A center field vector can distinguish the stated uniform ring and semicircle under the same geometry and calibration assumptions. Extra potential samples away from the center can also help, but one must derive their geometry rather than reuse an axis formula for an off-axis arc.

Worked original case

Investigation

Changed-case practice

Try a new practice case

Keep Q fixed and double the ring radius. What is the new center potential divided by its original value?

Numerical result in the requested units

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 evidence boundary

Original teaching and cases checked 30 October 2026 against the current official framework, Unit 9 and OpenStax electric potential. All records are supplied or generated, not measured by a learner. No physical charging, battery connection, high-voltage equipment or particle experiment is requested. Retain predictions, derivations and critique in the course evidence workspace; public feedback is not independent mastery or performed laboratory time.

Apply

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

Which reference, source geometry or evidence-independence assumption limits your inference?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Normalize a line potential and recover its field Next: Match potential branches without a false infinity reference

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