University Physics II · topic UPII-03.02
Gauss's law in integral form
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.
Gauss's law in integral form: predict a finite change
Gauss's law says the net electric flux out of any closed surface equals the net charge enclosed by that surface divided by ε₀, whatever the surface's shape and whatever charges sit outside it. Use the supplied topic model Y_out = 89.88*(qin/(rad_*rad_)+qout/((11+rad_)*(11+rad_))), where Y_out is FAR-SIDE E_r (kN C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. rad_ is Gaussian radius, measured in cm. Keep qin (Enclosed charge q_in) = 3 nC; qout (Outside charge q_out) = 4 nC fixed. For this exercise use only 2.925 cm ≤ rad_ ≤ 6.075 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 rad_ = 3.618 cm and 4.059 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrad_ across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FAR-SIDE E_r against rad_
Predictions from the supplied Gauss's law in integral form 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): FAR-SIDE E_r / kN C⁻¹
Horizontal axis (x): rad_: Gaussian radius / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rad_: Gaussian radius / cm | FAR-SIDE E_r / kN C⁻¹ |
|---|---|
| 2.925 | 33.370205661343526 |
| 3.7124999999999995 | 21.224643298843247 |
| 4.5 | 14.811996762631518 |
| 5.2875 | 10.999818998354925 |
| 6.074999999999999 | 8.53931278505241 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rad_ separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rad_ = 3.618 cm, substitute into Y_out = 89.88*(qin/(rad_*rad_)+qout/((11+rad_)*(11+rad_))): Y₀ = 22.2815 kN C⁻¹.
- At rad_ = 4.059 cm, the same fixed-parameter model gives Y₁ = 17.9515 kN C⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -4.33001 kN C⁻¹; the input change is Δrad_ = 0.441 cm.
- Average sensitivity is ΔY_out/Δrad_ = -9.81862 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 = 22.2815 kN C⁻¹, Y₁ = 17.9515 kN C⁻¹; ΔY_out = -4.33001 kN C⁻¹; average sensitivity = -9.81862 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.
Gauss's law in integral form: infer an input with uncertainty
Gauss's law says the net electric flux out of any closed surface equals the net charge enclosed by that surface divided by ε₀, whatever the surface's shape and whatever charges sit outside it. Use the supplied topic model Y_out = 89.88*(qin/(rad_*rad_)+qout/((11+rad_)*(11+rad_))), where Y_out is FAR-SIDE E_r (kN C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. rad_ is Gaussian radius, measured in cm. Keep qin (Enclosed charge q_in) = 3 nC; qout (Outside charge q_out) = 4 nC fixed. For this exercise use only 2.925 cm ≤ rad_ ≤ 6.075 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 = 12.911344 kN C⁻¹, with a bounded reading uncertainty ±0.6558 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 rad_ by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315 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.
FAR-SIDE E_r against rad_
Predictions from the supplied Gauss's law in integral form 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): FAR-SIDE E_r / kN C⁻¹
Horizontal axis (x): rad_: Gaussian radius / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rad_: Gaussian radius / cm | FAR-SIDE E_r / kN C⁻¹ |
|---|---|
| 2.925 | 33.370205661343526 |
| 3.7124999999999995 | 21.224643298843247 |
| 4.5 | 14.811996762631518 |
| 5.2875 | 10.999818998354925 |
| 6.074999999999999 | 8.53931278505241 |
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(rad_) = 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(2.925) = 33.3702 kN C⁻¹ and Y_out(6.075) = 8.53931 kN C⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rad__line = 2.925 + (12.9113 − 33.3702)(6.075 − 2.925)/(8.53931 − 33.3702) = 5.52037 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 rad_ = 4.8465 cm; substituting back gives Y_out = 12.911344 kN C⁻¹.
- The interpolation error in the input is |rad__line − rad_| = 0.673872 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 12.255544 kN C⁻¹ and 13.567144 kN C⁻¹. Sort the roots even if the response decreases: 4.71829 cm ≤ rad_ ≤ 4.98546 cm.
- The lower and upper input deviations are 0.12821 cm and 0.138958 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 rad_ ≈ 4.8465 cm; compatible interval [4.71829, 4.98546] cm. Endpoint interpolation gives 5.52037 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.