SCIENCE · INVESTIGATE AND EXPLAIN

Read a field from a potential landscape

Use spatial derivatives and equipotential geometry rather than a potential value alone.

Learning outcome: Compare C = −4, 0 and +4 V at fixed A. Retain the same contour shapes with relabeled potential values and the unchanged field. Then change A, calculate a local normal and tangent at (1,1), and predict which contour shape changes.

Starting recap

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

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Try

With A = 2 and C = 0, V(1,1) = 7 V and E(1,1) = (−6,−8) N/C. Adding C = 4 V changes the potential to 11 V but leaves E unchanged. A negative elementary charge moving from the origin to (1,1) changes potential energy by −7 eV.

Open the experiment

Predict

An additive constant is added to a static potential everywhere. What changes?

Your prediction and reason

Experiment

Compare C = −4, 0 and +4 V at fixed A. Retain the same contour shapes with relabeled potential values and the unchanged field. Then change A, calculate a local normal and tangent at (1,1), and predict which contour shape changes.

Differentiate a gauge-shifted spatial potential

Worked teaching model · synthetic values · Assigned static potential V = A x² + Bxy + D y² + C, with B = 2 V/m² and D = 3 V/m². Evaluate at (1,1) m. Positive-definite contours for supported A. This local polynomial is not a globally isolated finite-charge potential.

Quadratic x coefficient A: 2 V/m² Additive potential reference C: 0 V

Reset keeps your prediction, captures and practice history.

Equipotentials and field normals on equal spatial scales

Both position axes use 65 drawing units per metre; field arrows use 3 drawing units per N/C. Contours have V − C = 2, 5 and 10 V. Changing C relabels them without changing shape or arrows. These are local field directions, not particle trajectories.

x (m)	y (m)	Ex (N/C)	Ey (N/C)
1	1	-6	-8
-1	1	2	-4
1	-1	-2	4
-1	-1	6	8
0	1	-2	-6

Contour coordinate alternatives

V = 2 V

x (m)	y (m)
1	0
0.752738	0.342282
0.390879	0.632456
-0.0304881	0.826343
-0.447214	0.894427
-0.795855	0.826343
-1.02333	0.632456
-1.09502	0.342282
-1	1.09536e-16
-0.752738	-0.342282
-0.390879	-0.632456
0.0304881	-0.826343
0.447214	-0.894427
0.795855	-0.826343
1.02333	-0.632456
1.09502	-0.342282
1	-2.19071e-16

V = 5 V

x (m)	y (m)
1.58114	0
1.19018	0.541196
0.618034	1
-0.0482058	1.30656
-0.707107	1.41421
-1.25836	1.30656
-1.61803	1
-1.73138	0.541196
-1.58114	1.73191e-16
-1.19018	-0.541196
-0.618034	-1
0.0482058	-1.30656
0.707107	-1.41421
1.25836	-1.30656
1.61803	-1
1.73138	-0.541196
1.58114	-3.46382e-16

V = 10 V

x (m)	y (m)
2.23607	0
1.68317	0.765367
0.874032	1.41421
-0.0681734	1.84776
-1	2
-1.77959	1.84776
-2.28825	1.41421
-2.44854	0.765367
-2.23607	2.44929e-16
-1.68317	-0.765367
-0.874032	-1.41421
0.0681734	-1.84776
1	-2
1.77959	-1.84776
2.28825	-1.41421
2.44854	-0.765367
2.23607	-4.89859e-16

Potential along y = 1 m

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.

Potential along y = 1 m: numerical coordinates

x (m)	V(x,1) (V)
-2	7
-1.93333	6.60889
-1.86667	6.23556
-1.8	5.88
-1.73333	5.54222
-1.66667	5.22222
-1.6	4.92
-1.53333	4.63556
-1.46667	4.36889
-1.4	4.12
-1.33333	3.88889
-1.26667	3.67556
-1.2	3.48
-1.13333	3.30222
-1.06667	3.14222
-1	3
-0.933333	2.87556
-0.866667	2.76889
-0.8	2.68
-0.733333	2.60889
-0.666667	2.55556
-0.6	2.52
-0.533333	2.50222
-0.466667	2.50222
-0.4	2.52
-0.333333	2.55556
-0.266667	2.60889
-0.2	2.68
-0.133333	2.76889
-0.0666667	2.87556
0	3
0.0666667	3.14222
0.133333	3.30222
0.2	3.48
0.266667	3.67556
0.333333	3.88889
0.4	4.12
0.466667	4.36889
0.533333	4.63556
0.6	4.92
0.666667	5.22222
0.733333	5.54222
0.8	5.88
0.866667	6.23556
0.933333	6.60889
1	7
1.06667	7.40889
1.13333	7.83556
1.2	8.28
1.26667	8.74222
1.33333	9.22222
1.4	9.72
1.46667	10.2356
1.53333	10.7689
1.6	11.32
1.66667	11.8889
1.73333	12.4756
1.8	13.08
1.86667	13.7022
1.93333	14.3422
2	15

Field component along y = 1 m

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.

Field component along y = 1 m: numerical coordinates

x (m)	Ex(x,1) (N/C)
-2	6
-1.93333	5.73333
-1.86667	5.46667
-1.8	5.2
-1.73333	4.93333
-1.66667	4.66667
-1.6	4.4
-1.53333	4.13333
-1.46667	3.86667
-1.4	3.6
-1.33333	3.33333
-1.26667	3.06667
-1.2	2.8
-1.13333	2.53333
-1.06667	2.26667
-1	2
-0.933333	1.73333
-0.866667	1.46667
-0.8	1.2
-0.733333	0.933333
-0.666667	0.666667
-0.6	0.4
-0.533333	0.133333
-0.466667	-0.133333
-0.4	-0.4
-0.333333	-0.666667
-0.266667	-0.933333
-0.2	-1.2
-0.133333	-1.46667
-0.0666667	-1.73333
0	-2
0.0666667	-2.26667
0.133333	-2.53333
0.2	-2.8
0.266667	-3.06667
0.333333	-3.33333
0.4	-3.6
0.466667	-3.86667
0.533333	-4.13333
0.6	-4.4
0.666667	-4.66667
0.733333	-4.93333
0.8	-5.2
0.866667	-5.46667
0.933333	-5.73333
1	-6
1.06667	-6.26667
1.13333	-6.53333
1.2	-6.8
1.26667	-7.06667
1.33333	-7.33333
1.4	-7.6
1.46667	-7.86667
1.53333	-8.13333
1.6	-8.4
1.66667	-8.66667
1.73333	-8.93333
1.8	-9.2
1.86667	-9.46667
1.93333	-9.73333
2	-10

Calculated values under the declared model
Quantity	Value	Unit
Potential at (1,1)	7	V
Field x component at (1,1)	-6	N/C
Field y component at (1,1)	-8	N/C
Potential difference from (0,0) to (1,1)	7	V
Potential energy change for a negative elementary charge	-7	eV
Potential at origin	0	V

An additive reference changes displayed potential but not its spatial derivative or any potential difference.

Ex = −partial V/partial x and Ey = −partial V/partial y. A negative charge has force opposite the field and a potential-energy change opposite the sign of deltaV. Field vectors point toward decreasing V, not necessarily toward decreasing U for every charge sign.

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

An additive reference changes displayed potential but not its spatial derivative or any potential difference.

Observation to attach to a capture

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

Understand

Specify a local scalar landscape

The assigned potential is V(x,y) = A x² + 2xy + 3y² + C in SI units, with dimensional coefficients understood. It is a local static mathematical packet, not a claim about an isolated finite source extending to infinity. The field is E = −grad V: Ex = −2Ax − 2y and Ey = −2x − 6y.

Separate gauge from gradient

Changing C shifts every potential by the same constant. It does not alter a derivative, a potential difference, work or force. Potential energy for a chosen test charge is U = qV; a negative charge’s force points opposite E, although E itself always points toward decreasing V.

Construct and test an equipotential

Along a contour V = constant, dV = grad V dot dr = 0. Therefore the electrostatic field is perpendicular to the contour and has no tangential component. At A = 2, the gradient at (1,1) is (6,8) V/m. A tangent displacement proportional to (4,−3) has dot product 24 − 24 = 0 to first order. A finite straight tangent segment need not stay on the curved contour: distinguish a local tangent from an entire equipotential path.

Potential per charge and a chemical source

For a passive test charge, V = U/q and a difference is delta V = delta U/q; source potential does not depend on which test charge samples it. One volt is one joule per coulomb. A chemical cell can use non-electrostatic chemical processes to maintain separated charge and a terminal potential difference. In a supplied ideal packet, moving 0.4 C through a maintained 5 V difference transfers 2 J. A second cell moving 0.8 C through the same difference transfers 4 J: equal voltage is not equal available energy. This is an energy account, not a request to connect or handle a cell.

Worked original case

Investigation

Changed-case practice

Try a new practice case

Set A = 3 and C = −2. What is the x component of E at (1,1), in N/C?

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

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