Normalize a line potential and recover its field
Integrate scalar source contributions, then differentiate with respect to the observer.
Learning outcome: Compare beta = 0, 1 and 2 at the same total charge. Predict how moving more charge toward the near endpoint changes potential. Derive the field separately, compare the centered derivative discrepancy and test a changed perpendicular-bisector observer using the written expression.
Starting recap
Keep source normalization, potential reference, signed energy and evidence independence explicit.
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Try
For beta = 0 and z = 2 m, lambda0 = 2 nC/m and V = k(2 nC/m) ln 2 ≈ 12.4594 V. The field is k(2 nC/m)(1 − 1/2) = 8.98755 N/C. A centered finite difference of V approximates this derivative but has truncation error.
Open the experimentPredict
When differentiating the potential with respect to observer position, what must remain fixed?
Experiment
Compare beta = 0, 1 and 2 at the same total charge. Predict how moving more charge toward the near endpoint changes potential. Derive the field separately, compare the centered derivative discrepancy and test a changed perpendicular-bisector observer using the written expression.
Integrate line potential and recover its field
Worked teaching model · synthetic values · Finite source 0≤x≤1 m, Q = 2 nC and normalized lambda = lambda0(1 + beta x/L), L = 1 m. V(infinity) = 0 is valid for this localized source. Observer remains beyond source endpoint; no singular source-point potential.
Reset keeps your prediction, captures and practice history.
Potential beyond finite line
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 beyond finite line: numerical coordinates
| observer z (m) | V (V) |
|---|---|
| 1.5 | 20.9294 |
| 1.55833 | 19.484 |
| 1.61667 | 18.2378 |
| 1.675 | 17.1505 |
| 1.73333 | 16.1922 |
| 1.79167 | 15.3403 |
| 1.85 | 14.5773 |
| 1.90833 | 13.8896 |
| 1.96667 | 13.2662 |
| 2.025 | 12.6983 |
| 2.08333 | 12.1784 |
| 2.14167 | 11.7007 |
| 2.2 | 11.26 |
| 2.25833 | 10.8522 |
| 2.31667 | 10.4735 |
| 2.375 | 10.121 |
| 2.43333 | 9.79192 |
| 2.49167 | 9.48398 |
| 2.55 | 9.19518 |
| 2.60833 | 8.92375 |
| 2.66667 | 8.66815 |
| 2.725 | 8.42702 |
| 2.78333 | 8.19913 |
| 2.84167 | 7.98342 |
| 2.9 | 7.77893 |
| 2.95833 | 7.58478 |
| 3.01667 | 7.4002 |
| 3.075 | 7.2245 |
| 3.13333 | 7.05705 |
| 3.19167 | 6.89726 |
| 3.25 | 6.74462 |
| 3.30833 | 6.59866 |
| 3.36667 | 6.45894 |
| 3.425 | 6.32507 |
| 3.48333 | 6.19669 |
| 3.54167 | 6.07346 |
| 3.6 | 5.95507 |
| 3.65833 | 5.84124 |
| 3.71667 | 5.73172 |
| 3.775 | 5.62626 |
| 3.83333 | 5.52464 |
| 3.89167 | 5.42665 |
| 3.95 | 5.33209 |
| 4.00833 | 5.2408 |
| 4.06667 | 5.1526 |
| 4.125 | 5.06733 |
| 4.18333 | 4.98486 |
| 4.24167 | 4.90505 |
| 4.3 | 4.82776 |
| 4.35833 | 4.75288 |
| 4.41667 | 4.68031 |
| 4.475 | 4.60993 |
| 4.53333 | 4.54164 |
| 4.59167 | 4.47535 |
| 4.65 | 4.41099 |
| 4.70833 | 4.34845 |
| 4.76667 | 4.28767 |
| 4.825 | 4.22858 |
| 4.88333 | 4.17109 |
| 4.94167 | 4.11516 |
| 5 | 4.06071 |
| Quantity | Value | Unit |
|---|---|---|
| Normalized base density | 1.33333 | nC/m |
| Axial potential | 12.9354 | V |
| Analytic negative potential derivative | 9.66884 | N/C |
| Centered derivative check | 9.66884 | N/C |
| Derivative check signed discrepancy | 5.99419e-8 | N/C |
Scalar source integration and a spatial derivative reproduce the independently derived axial field.
Use 1 + beta x = (1 + beta z) − beta(z−x) inside the potential integral. The result is k lambda0[(1 + beta z)ln(z/(z−1)) − beta] for the declared 1 m source. The finite-difference derivative has truncation error; it is not exact or independently measured.
Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.
Scalar source integration and a spatial derivative reproduce the independently derived axial field.
Use 1 + beta x = (1 + beta z) − beta(z−x) inside the potential integral. The result is k lambda0[(1 + beta z)ln(z/(z−1)) − beta] for the declared 1 m source. The finite-difference derivative has truncation error; it is not exact or independently measured.
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