University Physics II · topic UPII-01.07
Electrostatic material response
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.
Electrostatic material response: predict a finite change
Electrostatic material response is what a neutral material does in an applied field: free charge moves to the surfaces, bound charge stretches into induced dipoles, and both leave the near face oppositely charged to the source. Use the supplied topic model Y_out = 0.1798*q^2/r^5, where Y_out is INDUCED PULL F (μN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Centre-to-centre separation r, measured in cm. Keep s (Sign of the source charge) = 1; q (Source charge magnitude |q|) = 30 nC fixed. For this exercise use only 1.65938 cm ≤ r ≤ 2.84063 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 = 1.91925 cm and 2.08462 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.
INDUCED PULL F against r
Predictions from the supplied Electrostatic material response 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): INDUCED PULL F / μN
Horizontal axis (x): r: Centre-to-centre separation r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Centre-to-centre separation r / cm | INDUCED PULL F / μN |
|---|---|
| 1.659375 | 12.862028957876241 |
| 1.9546875000000001 | 5.670815507120873 |
| 2.25 | 2.8062063709800333 |
| 2.5453125 | 1.514700069050953 |
| 2.840625 | 0.8749085734767675 |
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 = 1.91925 cm, substitute into Y_out = 0.1798*q^2/r^5: Y₀ = 6.21405 μN.
- At r = 2.08462 cm, the same fixed-parameter model gives Y₁ = 4.11048 μN.
- Signed change: ΔY_out = Y₁ − Y₀ = -2.10357 μN; the input change is Δr = 0.165375 cm.
- Average sensitivity is ΔY_out/Δr = -12.72 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 = 6.21405 μN, Y₁ = 4.11048 μN; ΔY_out = -2.10357 μN; average sensitivity = -12.72 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.
Electrostatic material response: infer an input with uncertainty
Electrostatic material response is what a neutral material does in an applied field: free charge moves to the surfaces, bound charge stretches into induced dipoles, and both leave the near face oppositely charged to the source. Use the supplied topic model Y_out = 0.1798*q^2/r^5, where Y_out is INDUCED PULL F (μN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Centre-to-centre separation r, measured in cm. Keep s (Sign of the source charge) = 1; q (Source charge magnitude |q|) = 30 nC fixed. For this exercise use only 1.65938 cm ≤ r ≤ 2.84063 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 = 2.1193586 μN, with a bounded reading uncertainty ±0.18667 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 0.000118125 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.
INDUCED PULL F against r
Predictions from the supplied Electrostatic material response 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): INDUCED PULL F / μN
Horizontal axis (x): r: Centre-to-centre separation r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Centre-to-centre separation r / cm | INDUCED PULL F / μN |
|---|---|
| 1.659375 | 12.862028957876241 |
| 1.9546875000000001 | 5.670815507120873 |
| 2.25 | 2.8062063709800333 |
| 2.5453125 | 1.514700069050953 |
| 2.840625 | 0.8749085734767675 |
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(1.65938) = 12.862 μN and Y_out(2.84063) = 0.874909 μN. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 1.65938 + (2.11936 − 12.862)(2.84063 − 1.65938)/(0.874909 − 12.862) = 2.71799 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 = 2.37994 cm; substituting back gives Y_out = 2.1193586 μN.
- The interpolation error in the input is |r_line − r| = 0.338055 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.9326886 μN and 2.3060286 μN. Sort the roots even if the response decreases: 2.3401 cm ≤ r ≤ 2.42423 cm.
- The lower and upper input deviations are 0.0398424 cm and 0.0442939 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 ≈ 2.37994 cm; compatible interval [2.3401, 2.42423] cm. Endpoint interpolation gives 2.71799 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.