University Physics II · topic UPII-01.04
Coulomb's law
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Coulomb's Law: predict a finite change
Coulomb's law says that two point charges at rest push or pull each other along the line joining them, with a force proportional to the product of their charge magnitudes and inversely proportional to the square of their separation. Use the supplied topic model Y_out = 89.88*abs(q1*q2)/(r*r), where Y_out is FORCE ON EACH F (µN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|. r is Separation r, measured in cm. Keep q1 (Charge q₁ at the left) = 5 nC; q2 (Charge q₂ on the right) = -3 nC fixed. For this exercise use only 4.6375 cm ≤ r ≤ 9.3625 cm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at r = 5.677 cm and 6.3385 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FORCE ON EACH F against r
Predictions from the supplied Coulomb's Law model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): FORCE ON EACH F / µN
Horizontal axis (x): r: Separation r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Separation r / cm | FORCE ON EACH F / µN |
|---|---|
| 4.6375 | 62.68829781823729 |
| 5.8187500000000005 | 39.81941965547595 |
| 7 | 27.51428571428571 |
| 8.18125 | 20.14257526053688 |
| 9.3625 | 15.38050734312416 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = 5.677 cm, substitute into Y_out = 89.88*abs(q1*q2)/(r*r): Y₀ = 41.8328 µN.
- At r = 6.3385 cm, the same fixed-parameter model gives Y₁ = 33.5569 µN.
- Signed change: ΔY_out = Y₁ − Y₀ = -8.2759 µN; the input change is Δr = 0.6615 cm.
- Average sensitivity is ΔY_out/Δr = -12.5108 µN per cm. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 41.8328 µN, Y₁ = 33.5569 µN; ΔY_out = -8.2759 µN; average sensitivity = -12.5108 µN per cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Coulomb's Law: infer an input with uncertainty
Coulomb's law says that two point charges at rest push or pull each other along the line joining them, with a force proportional to the product of their charge magnitudes and inversely proportional to the square of their separation. Use the supplied topic model Y_out = 89.88*abs(q1*q2)/(r*r), where Y_out is FORCE ON EACH F (µN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|. r is Separation r, measured in cm. Keep q1 (Charge q₁ at the left) = 5 nC; q2 (Charge q₂ on the right) = -3 nC fixed. For this exercise use only 4.6375 cm ≤ r ≤ 9.3625 cm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 23.842265 µN, with a bounded reading uncertainty ±1.2693 µN. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 cm, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
FORCE ON EACH F against r
Predictions from the supplied Coulomb's Law model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): FORCE ON EACH F / µN
Horizontal axis (x): r: Separation r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Separation r / cm | FORCE ON EACH F / µN |
|---|---|
| 4.6375 | 62.68829781823729 |
| 5.8187500000000005 | 39.81941965547595 |
| 7 | 27.51428571428571 |
| 8.18125 | 20.14257526053688 |
| 9.3625 | 15.38050734312416 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(r) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(4.6375) = 62.6883 µN and Y_out(9.3625) = 15.3805 µN. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 4.6375 + (23.8423 − 62.6883)(9.3625 − 4.6375)/(15.3805 − 62.6883) = 8.51736 cm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives r = 7.51975 cm; substituting back gives Y_out = 23.842265 µN.
- The interpolation error in the input is |r_line − r| = 0.997608 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 22.572965 µN and 25.111565 µN. Sort the roots even if the response decreases: 7.32724 cm ≤ r ≤ 7.72828 cm.
- The lower and upper input deviations are 0.192513 cm and 0.20853 cm. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer r ≈ 7.51975 cm; compatible interval [7.32724, 7.72828] cm. Endpoint interpolation gives 8.51736 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.