University Physics II · topic UPII-03.04
Spherical charge distributions
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.
Spherical charge distributions: predict a finite change
A spherically symmetric charge distribution is one whose charge density depends only on the distance r from a single centre; its field is radial everywhere, and at radius r it is fixed entirely by the charge enclosed within r. Use the supplied topic model Y_out = (539.28/rs^2)*clamp(((rp/rs)^3-c^3)/(1-c^3),0,1)/(rp/rs)^2, where Y_out is FIELD AT PROBE (kV m⁻¹). 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]. rp is Probe radius r, measured in cm. Keep rs (Outer radius R) = 6 cm; c (Empty core fraction a/R) = 0 fixed. For this exercise use only 6.66687 cm ≤ rp ≤ 17.9281 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 rp = 9.14435 cm and 10.7209 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FIELD AT PROBE against rp
Predictions from the supplied Spherical charge distributions 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 PROBE / kV m⁻¹
Horizontal axis (x): rp: Probe radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rp: Probe radius r / cm | FIELD AT PROBE / kV m⁻¹ |
|---|---|
| 6.666874999999999 | 12.13304167304676 |
| 9.482187499999998 | 5.9978725980215915 |
| 12.2975 | 3.5659943576279294 |
| 15.112812499999999 | 2.3611508030536323 |
| 17.928124999999998 | 1.6778169248910673 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rp separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rp = 9.14435 cm, substitute into Y_out = (539.28/rs^2)*clamp(((rp/rs)^3-c^3)/(1-c^3),0,1)/(rp/rs)^2: Y₀ = 6.44924 kV m⁻¹.
- At rp = 10.7209 cm, the same fixed-parameter model gives Y₁ = 4.69191 kV m⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.75733 kV m⁻¹; the input change is Δrp = 1.57657 cm.
- Average sensitivity is ΔY_out/Δrp = -1.11465 kV m⁻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 = 6.44924 kV m⁻¹, Y₁ = 4.69191 kV m⁻¹; ΔY_out = -1.75733 kV m⁻¹; average sensitivity = -1.11465 kV m⁻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.
Spherical charge distributions: infer an input with uncertainty
A spherically symmetric charge distribution is one whose charge density depends only on the distance r from a single centre; its field is radial everywhere, and at radius r it is fixed entirely by the charge enclosed within r. Use the supplied topic model Y_out = (539.28/rs^2)*clamp(((rp/rs)^3-c^3)/(1-c^3),0,1)/(rp/rs)^2, where Y_out is FIELD AT PROBE (kV m⁻¹). 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]. rp is Probe radius r, measured in cm. Keep rs (Outer radius R) = 6 cm; c (Empty core fraction a/R) = 0 fixed. For this exercise use only 6.66687 cm ≤ rp ≤ 17.9281 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 = 2.9431906 kV m⁻¹, with a bounded reading uncertainty ±0.18981 kV m⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00112612 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 PROBE against rp
Predictions from the supplied Spherical charge distributions 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 PROBE / kV m⁻¹
Horizontal axis (x): rp: Probe radius r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rp: Probe radius r / cm | FIELD AT PROBE / kV m⁻¹ |
|---|---|
| 6.666874999999999 | 12.13304167304676 |
| 9.482187499999998 | 5.9978725980215915 |
| 12.2975 | 3.5659943576279294 |
| 15.112812499999999 | 2.3611508030536323 |
| 17.928124999999998 | 1.6778169248910673 |
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(rp) = 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(6.66687) = 12.133 kV m⁻¹ and Y_out(17.9281) = 1.67782 kV m⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rp_line = 6.66687 + (2.94319 − 12.133)(17.9281 − 6.66687)/(1.67782 − 12.133) = 16.5652 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 rp = 13.5362 cm; substituting back gives Y_out = 2.9431906 kV m⁻¹.
- The interpolation error in the input is |rp_line − rp| = 3.02896 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.7533806 kV m⁻¹ and 3.1330006 kV m⁻¹. Sort the roots even if the response decreases: 13.1198 cm ≤ rp ≤ 13.995 cm.
- The lower and upper input deviations are 0.416446 cm and 0.458799 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 rp ≈ 13.5362 cm; compatible interval [13.1198, 13.995] cm. Endpoint interpolation gives 16.5652 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.