University Physics II · topic UPII-03.07
Gauss's law and conductors
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 and conductors: predict a finite change
A conductor in electrostatic equilibrium carries no electric field inside its conducting material, so all of its net charge lies on surfaces, and the field just outside is perpendicular to the surface with magnitude σ/ε₀. Use the supplied topic model Y_out = 89.88*q/a^2, where Y_out is E AT THE WALL (kV m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. a is Inner radius a, measured in cm. Keep q (Cavity charge q) = 3 nC; qs (Shell net charge Q) = 2 nC fixed. For this exercise use only 8.6375 cm ≤ a ≤ 13.3625 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 a = 9.677 cm and 10.3385 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
E AT THE WALL against a
Predictions from the supplied Gauss's law and conductors 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 AT THE WALL / kV m⁻¹
Horizontal axis (x): a: Inner radius a / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Inner radius a / cm | E AT THE WALL / kV m⁻¹ |
|---|---|
| 8.6375 | 3.614166846429492 |
| 9.81875 | 2.7968676371259633 |
| 11 | 2.2284297520661154 |
| 12.18125 | 1.8171911185196865 |
| 13.3625 | 1.5101110380910792 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 9.677 cm, substitute into Y_out = 89.88*q/a^2: Y₀ = 2.87941 kV m⁻¹.
- At a = 10.3385 cm, the same fixed-parameter model gives Y₁ = 2.52272 kV m⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.356684 kV m⁻¹; the input change is Δa = 0.6615 cm.
- Average sensitivity is ΔY_out/Δa = -0.539205 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 = 2.87941 kV m⁻¹, Y₁ = 2.52272 kV m⁻¹; ΔY_out = -0.356684 kV m⁻¹; average sensitivity = -0.539205 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.
Gauss's law and conductors: infer an input with uncertainty
A conductor in electrostatic equilibrium carries no electric field inside its conducting material, so all of its net charge lies on surfaces, and the field just outside is perpendicular to the surface with magnitude σ/ε₀. Use the supplied topic model Y_out = 89.88*q/a^2, where Y_out is E AT THE WALL (kV m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. a is Inner radius a, measured in cm. Keep q (Cavity charge q) = 3 nC; qs (Shell net charge Q) = 2 nC fixed. For this exercise use only 8.6375 cm ≤ a ≤ 13.3625 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 = 2.0318807 kV m⁻¹, with a bounded reading uncertainty ±0.078265 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 a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 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 AT THE WALL against a
Predictions from the supplied Gauss's law and conductors 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 AT THE WALL / kV m⁻¹
Horizontal axis (x): a: Inner radius a / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Inner radius a / cm | E AT THE WALL / kV m⁻¹ |
|---|---|
| 8.6375 | 3.614166846429492 |
| 9.81875 | 2.7968676371259633 |
| 11 | 2.2284297520661154 |
| 12.18125 | 1.8171911185196865 |
| 13.3625 | 1.5101110380910792 |
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(a) = 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(8.6375) = 3.61417 kV m⁻¹ and Y_out(13.3625) = 1.51011 kV m⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 8.6375 + (2.03188 − 3.61417)(13.3625 − 8.6375)/(1.51011 − 3.61417) = 12.1908 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 a = 11.5198 cm; substituting back gives Y_out = 2.0318807 kV m⁻¹.
- The interpolation error in the input is |a_line − a| = 0.671031 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.9536157 kV m⁻¹ and 2.1101457 kV m⁻¹. Sort the roots even if the response decreases: 11.3041 cm ≤ a ≤ 11.7482 cm.
- The lower and upper input deviations are 0.215651 cm and 0.228484 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 a ≈ 11.5198 cm; compatible interval [11.3041, 11.7482] cm. Endpoint interpolation gives 12.1908 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.