University Physics II · topic UPII-03.05
Cylindrical symmetry
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.
Cylindrical symmetry: predict a finite change
A distribution has cylindrical symmetry when rotating it about one axis, sliding it along that axis, reflecting it in a plane perpendicular to that axis, or reflecting it in a plane containing that axis all leave it unchanged — which forces its field to lie along the radius, outward for positive charge and inward for negative, with a magnitude set by the distance from the axis alone. Use the supplied topic model Y_out = 0.56469*rho*min(p, rc*rc/p), where Y_out is FIELD AT THE PROBE (kN C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. p is Probe distance r, measured in cm. Keep rc (Cylinder radius R) = 2 cm; rho (Charge density ρ) = 4 μC m⁻³ fixed. For this exercise use only 2.28125 cm ≤ p ≤ 7.20312 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 p = 3.36406 cm and 4.05312 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FIELD AT THE PROBE against p
Predictions from the supplied Cylindrical symmetry 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): FIELD AT THE PROBE / kN C⁻¹
Horizontal axis (x): p: Probe distance r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Probe distance r / cm | FIELD AT THE PROBE / kN C⁻¹ |
|---|---|
| 2.28125 | 3.960565479452055 |
| 3.51171875 | 2.5728256284760844 |
| 4.7421875 | 1.9052473146622735 |
| 5.972656249999999 | 1.5127339699149773 |
| 7.203124999999999 | 1.254322255965293 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p = 3.36406 cm, substitute into Y_out = 0.56469*rho*min(p, rc*rc/p): Y₀ = 2.68575 kN C⁻¹.
- At p = 4.05312 cm, the same fixed-parameter model gives Y₁ = 2.22915 kN C⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.456599 kN C⁻¹; the input change is Δp = 0.689062 cm.
- Average sensitivity is ΔY_out/Δp = -0.662638 kN C⁻¹ 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 = 2.68575 kN C⁻¹, Y₁ = 2.22915 kN C⁻¹; ΔY_out = -0.456599 kN C⁻¹; average sensitivity = -0.662638 kN C⁻¹ 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.
Cylindrical symmetry: infer an input with uncertainty
A distribution has cylindrical symmetry when rotating it about one axis, sliding it along that axis, reflecting it in a plane perpendicular to that axis, or reflecting it in a plane containing that axis all leave it unchanged — which forces its field to lie along the radius, outward for positive charge and inward for negative, with a magnitude set by the distance from the axis alone. Use the supplied topic model Y_out = 0.56469*rho*min(p, rc*rc/p), where Y_out is FIELD AT THE PROBE (kN C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. p is Probe distance r, measured in cm. Keep rc (Cylinder radius R) = 2 cm; rho (Charge density ρ) = 4 μC m⁻³ fixed. For this exercise use only 2.28125 cm ≤ p ≤ 7.20312 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 = 1.7100179 kN C⁻¹, with a bounded reading uncertainty ±0.068354 kN C⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000492187 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.
FIELD AT THE PROBE against p
Predictions from the supplied Cylindrical symmetry 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): FIELD AT THE PROBE / kN C⁻¹
Horizontal axis (x): p: Probe distance r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Probe distance r / cm | FIELD AT THE PROBE / kN C⁻¹ |
|---|---|
| 2.28125 | 3.960565479452055 |
| 3.51171875 | 2.5728256284760844 |
| 4.7421875 | 1.9052473146622735 |
| 5.972656249999999 | 1.5127339699149773 |
| 7.203124999999999 | 1.254322255965293 |
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(p) = 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(2.28125) = 3.96057 kN C⁻¹ and Y_out(7.20312) = 1.25432 kN C⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = 2.28125 + (1.71002 − 3.96057)(7.20312 − 2.28125)/(1.25432 − 3.96057) = 6.37435 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 p = 5.28359 cm; substituting back gives Y_out = 1.7100179 kN C⁻¹.
- The interpolation error in the input is |p_line − p| = 1.09075 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.6416639 kN C⁻¹ and 1.7783719 kN C⁻¹. Sort the roots even if the response decreases: 5.08051 cm ≤ p ≤ 5.50359 cm.
- The lower and upper input deviations are 0.203082 cm and 0.219993 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 p ≈ 5.28359 cm; compatible interval [5.08051, 5.50359] cm. Endpoint interpolation gives 6.37435 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.