University Physics II · topic UPII-03.03
Choosing a Gaussian surface
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.
Choosing a Gaussian surface: predict a finite change
A Gaussian surface is an imaginary closed surface chosen so that symmetry makes ∮E⋅dA trivial: on every piece, either E is constant in magnitude and perpendicular to the surface, or the flux through that piece is zero. Use the supplied topic model Y_out = 898.8*q - 89880*p/(d-10)^2, where Y_out is E⊥ NEAR SIDE (N C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is q₂ distance from centre, measured in cm. Keep q (Enclosed charge q₁) = 3 nC; p (Outside charge q₂) = 2 nC fixed. For this exercise use only 20.8687 cm ≤ d ≤ 42.1312 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 = 25.5465 cm and 28.5232 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.
E⊥ NEAR SIDE against d
Predictions from the supplied Choosing a Gaussian surface 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): E⊥ NEAR SIDE / N C⁻¹
Horizontal axis (x): d: q₂ 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: q₂ distance from centre / cm | E⊥ NEAR SIDE / N C⁻¹ |
|---|---|
| 20.86875 | 1174.6831110263106 |
| 26.184374999999996 | 2010.120234787501 |
| 31.499999999999996 | 2307.519524067063 |
| 36.815625 | 2446.413122413025 |
| 42.131249999999994 | 2522.2843465848364 |
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 = 25.5465 cm, substitute into Y_out = 898.8*q - 89880*p/(d-10)^2: Y₀ = 1952.65 N C⁻¹.
- At d = 28.5232 cm, the same fixed-parameter model gives Y₁ = 2172.49 N C⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 219.839 N C⁻¹; the input change is Δd = 2.97675 cm.
- Average sensitivity is ΔY_out/Δd = 73.852 N C⁻¹ per cm. Its positive sign means that this response increases 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 = 1952.65 N C⁻¹, Y₁ = 2172.49 N C⁻¹; ΔY_out = 219.839 N C⁻¹; average sensitivity = 73.852 N 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.
Choosing a Gaussian surface: infer an input with uncertainty
A Gaussian surface is an imaginary closed surface chosen so that symmetry makes ∮E⋅dA trivial: on every piece, either E is constant in magnitude and perpendicular to the surface, or the flux through that piece is zero. Use the supplied topic model Y_out = 898.8*q - 89880*p/(d-10)^2, where Y_out is E⊥ NEAR SIDE (N C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is q₂ distance from centre, measured in cm. Keep q (Enclosed charge q₁) = 3 nC; p (Outside charge q₂) = 2 nC fixed. For this exercise use only 20.8687 cm ≤ d ≤ 42.1312 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 = 2380.0837 N C⁻¹, with a bounded reading uncertainty ±21.33 N 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 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 0.00212625 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.
E⊥ NEAR SIDE against d
Predictions from the supplied Choosing a Gaussian surface 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): E⊥ NEAR SIDE / N C⁻¹
Horizontal axis (x): d: q₂ 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: q₂ distance from centre / cm | E⊥ NEAR SIDE / N C⁻¹ |
|---|---|
| 20.86875 | 1174.6831110263106 |
| 26.184374999999996 | 2010.120234787501 |
| 31.499999999999996 | 2307.519524067063 |
| 36.815625 | 2446.413122413025 |
| 42.131249999999994 | 2522.2843465848364 |
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(20.8687) = 1174.68 N C⁻¹ and Y_out(42.1312) = 2522.28 N C⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = 20.8687 + (2380.08 − 1174.68)(42.1312 − 20.8687)/(2522.28 − 1174.68) = 39.8876 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 = 33.8389 cm; substituting back gives Y_out = 2380.0837 N C⁻¹.
- The interpolation error in the input is |d_line − d| = 6.04873 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2358.7537 N C⁻¹ and 2401.4137 N C⁻¹. Sort the roots even if the response decreases: 33.0736 cm ≤ d ≤ 34.6857 cm.
- The lower and upper input deviations are 0.765265 cm and 0.846835 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 ≈ 33.8389 cm; compatible interval [33.0736, 34.6857] cm. Endpoint interpolation gives 39.8876 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.