University Physics II · topic UPII-01.03
Charging by friction, contact, and induction
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.
Charging by Friction, Contact & Induction: predict a finite change
Charging by friction, contact and induction are three ways of moving electrons between objects: rubbing transfers them at a surface, touching shares them between conductors, and induction uses a nearby charge plus a path to earth to strand a charge of the opposite sign. Use the supplied topic model Y_out = 4*q/d*clamp(stage-1,0,1), where Y_out is NET ON SPHERE (nC). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; clamp(x,a,b) confines x to [a,b]. d is Bead distance from centre, measured in cm. Keep stage (Step in the sequence) = 2; q (Bead charge magnitude) = 6 nC fixed. For this exercise use only 9.7 cm ≤ d ≤ 22.3 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 d = 12.472 cm and 14.236 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
NET ON SPHERE against d
Predictions from the supplied Charging by Friction, Contact & Induction 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): NET ON SPHERE / nC
Horizontal axis (x): d: Bead distance from centre / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Bead distance from centre / cm | NET ON SPHERE / nC |
|---|---|
| 9.7 | 2.4742268041237114 |
| 12.849999999999998 | 1.8677042801556423 |
| 15.999999999999998 | 1.5000000000000002 |
| 19.15 | 1.2532637075718016 |
| 22.299999999999997 | 1.0762331838565025 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 12.472 cm, substitute into Y_out = 4*q/d*clamp(stage-1,0,1): Y₀ = 1.92431 nC.
- At d = 14.236 cm, the same fixed-parameter model gives Y₁ = 1.68587 nC.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.238444 nC; the input change is Δd = 1.764 cm.
- Average sensitivity is ΔY_out/Δd = -0.135172 nC 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 = 1.92431 nC, Y₁ = 1.68587 nC; ΔY_out = -0.238444 nC; average sensitivity = -0.135172 nC 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.
Charging by Friction, Contact & Induction: infer an input with uncertainty
Charging by friction, contact and induction are three ways of moving electrons between objects: rubbing transfers them at a surface, touching shares them between conductors, and induction uses a nearby charge plus a path to earth to strand a charge of the opposite sign. Use the supplied topic model Y_out = 4*q/d*clamp(stage-1,0,1), where Y_out is NET ON SPHERE (nC). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; clamp(x,a,b) confines x to [a,b]. d is Bead distance from centre, measured in cm. Keep stage (Step in the sequence) = 2; q (Bead charge magnitude) = 6 nC fixed. For this exercise use only 9.7 cm ≤ d ≤ 22.3 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 = 1.380421 nC, with a bounded reading uncertainty ±0.045628 nC. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00126 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.
NET ON SPHERE against d
Predictions from the supplied Charging by Friction, Contact & Induction 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): NET ON SPHERE / nC
Horizontal axis (x): d: Bead distance from centre / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Bead distance from centre / cm | NET ON SPHERE / nC |
|---|---|
| 9.7 | 2.4742268041237114 |
| 12.849999999999998 | 1.8677042801556423 |
| 15.999999999999998 | 1.5000000000000002 |
| 19.15 | 1.2532637075718016 |
| 22.299999999999997 | 1.0762331838565025 |
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(d) = 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(9.7) = 2.47423 nC and Y_out(22.3) = 1.07623 nC. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 9.7 + (1.38042 − 2.47423)(22.3 − 9.7)/(1.07623 − 2.47423) = 19.5584 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 d = 17.386 cm; substituting back gives Y_out = 1.380421 nC.
- The interpolation error in the input is |d_line − d| = 2.17238 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.334793 nC and 1.426049 nC. Sort the roots even if the response decreases: 16.8297 cm ≤ d ≤ 17.9803 cm.
- The lower and upper input deviations are 0.556284 cm and 0.594316 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 d ≈ 17.386 cm; compatible interval [16.8297, 17.9803] cm. Endpoint interpolation gives 19.5584 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.