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University Physics II · topic UPII-05.06

Gauss's law in dielectric media

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.

Further practice · 01

Gauss's law in dielectric media: predict a finite change

In matter, Gauss's law still counts every enclosed charge, free plus bound. The displacement field D = ε₀E + P repackages that count so only free charge appears on the right: ∮D⋅dA = Q_free,enc. Use the supplied topic model Y_out = s-s/k, where Y_out is P = σ_b IN THE SLAB (μC m⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Dielectric constant κ, dimensionless. Keep s (Free charge density σ_f) = 1.77 μC m⁻²; p (Gap filled by the slab) = 55 % fixed. For this exercise use only 1.31875 ≤ k ≤ 3.68125 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 k = 1.8385 and 2.16925. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δk across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.

Explore the question data

P = σ_b IN THE SLAB against k

Predictions from the supplied Gauss's law in dielectric media 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): P = σ_b IN THE SLAB / μC m⁻²

P = σ_b IN THE SLAB against kPredictions from the supplied Gauss's law in dielectric media 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. Horizontal axis: k: Dielectric constant κ. Vertical axis: P = σ_b IN THE SLAB / μC m⁻². Exact values are available in the data table below.0.39340.62590.85851.0911.3241.2241.8622.53.1383.776xy

Horizontal axis (x): k: Dielectric constant κ

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1k: Dielectric constant κ: 1.319P = σ_b IN THE SLAB / μC m⁻²: 0.4278
Read the exact data as a table
P = σ_b IN THE SLAB against k — supplied values
k: Dielectric constant κP = σ_b IN THE SLAB / μC m⁻²
1.318750.4278199052132703
1.9093750.8429950900163666
2.51.062
3.0906251.1973003033367038
3.681251.2891850594227505

These are the supplied values, not a worked answer. No fitted trend is drawn.

Show a hint

Substitute each value of k separately while keeping every other input fixed, then subtract in the same order.

Reveal the worked solution
  1. At k = 1.8385, substitute into Y_out = s-s/k: Y₀ = 0.807259 μC m⁻².
  2. At k = 2.16925, the same fixed-parameter model gives Y₁ = 0.95405 μC m⁻².
  3. Signed change: ΔY_out = Y₁ − Y₀ = 0.146791 μC m⁻²; the input change is Δk = 0.33075.
  4. Average sensitivity is ΔY_out/Δk = 0.443813 μC m⁻² per unit input. Its positive sign means that this response increases over the specified interval.
  5. 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 = 0.807259 μC m⁻², Y₁ = 0.95405 μC m⁻²; ΔY_out = 0.146791 μC m⁻²; average sensitivity = 0.443813 μC m⁻² per unit input.

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.

Extra challenging · 01

Gauss's law in dielectric media: infer an input with uncertainty

In matter, Gauss's law still counts every enclosed charge, free plus bound. The displacement field D = ε₀E + P repackages that count so only free charge appears on the right: ∮D⋅dA = Q_free,enc. Use the supplied topic model Y_out = s-s/k, where Y_out is P = σ_b IN THE SLAB (μC m⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Dielectric constant κ, dimensionless. Keep s (Free charge density σ_f) = 1.77 μC m⁻²; p (Gap filled by the slab) = 55 % fixed. For this exercise use only 1.31875 ≤ k ≤ 3.68125 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 = 1.1286666 μC m⁻², with a bounded reading uncertainty ±0.024078 μC m⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625, 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.

Explore the question data

P = σ_b IN THE SLAB against k

Predictions from the supplied Gauss's law in dielectric media 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): P = σ_b IN THE SLAB / μC m⁻²

P = σ_b IN THE SLAB against kPredictions from the supplied Gauss's law in dielectric media 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. Horizontal axis: k: Dielectric constant κ. Vertical axis: P = σ_b IN THE SLAB / μC m⁻². Exact values are available in the data table below.0.39340.62590.85851.0911.3241.2241.8622.53.1383.776xy

Horizontal axis (x): k: Dielectric constant κ

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1k: Dielectric constant κ: 1.319P = σ_b IN THE SLAB / μC m⁻²: 0.4278
Read the exact data as a table
P = σ_b IN THE SLAB against k — supplied values
k: Dielectric constant κP = σ_b IN THE SLAB / μC m⁻²
1.318750.4278199052132703
1.9093750.8429950900163666
2.51.062
3.0906251.1973003033367038
3.681251.2891850594227505

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(k) = 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
  1. Endpoint predictions are Y_out(1.31875) = 0.42782 μC m⁻² and Y_out(3.68125) = 1.28919 μC m⁻². The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives k_line = 1.31875 + (1.12867 − 0.42782)(3.68125 − 1.31875)/(1.28919 − 0.42782) = 3.24099. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives k = 2.75987; substituting back gives Y_out = 1.1286666 μC m⁻².
  4. The interpolation error in the input is |k_line − k| = 0.481115. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 1.1045886 μC m⁻² and 1.1527446 μC m⁻². Sort the roots even if the response decreases: 2.66001 ≤ k ≤ 2.86753.
  6. The lower and upper input deviations are 0.0998664 and 0.107658. 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 k ≈ 2.75987; compatible interval [2.66001, 2.86753]. Endpoint interpolation gives 3.24099.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.