Skip to main content
Course and syllabus guidance

University Physics II · topic UPII-01.02

Conductors, insulators, and polarization

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

Conductors, insulators, and polarization: predict a finite change

A conductor carries charge that is free to travel through the whole sample, an insulator holds every charge bound to its own molecule, and polarization is the small separation of bound charge that a field produces inside neutral matter. Use the supplied topic model Y_out = 0.16689*nd*ef, where Y_out is POLARIZATION P = σ_b (μC m⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nd is Molecules per cubic metre n, measured in 10²⁷ m⁻³. Keep ef (Applied field E) = 3 MV m⁻¹ fixed. For this exercise use only 13.1875 10²⁷ m⁻³ ≤ nd ≤ 36.8125 10²⁷ m⁻³ 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 nd = 18.385 10²⁷ m⁻³ and 21.6925 10²⁷ m⁻³. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δnd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.

Explore the question data

POLARIZATION P = σ_b against nd

Predictions from the supplied Conductors, insulators, and polarization 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): POLARIZATION P = σ_b / μC m⁻²

POLARIZATION P = σ_b against ndPredictions from the supplied Conductors, insulators, and polarization 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: nd: Molecules per cubic metre n / 10²⁷ m⁻³. Vertical axis: POLARIZATION P = σ_b / μC m⁻². Exact values are available in the data table below.6.1299.32312.5215.7118.912.2418.622531.3837.76xy

Horizontal axis (x): nd: Molecules per cubic metre n / 10²⁷ m⁻³

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

Reading 1nd: Molecules per cubic metre n / 10²⁷ m⁻³: 13.19POLARIZATION P = σ_b / μC m⁻²: 6.603
Read the exact data as a table
POLARIZATION P = σ_b against nd — supplied values
nd: Molecules per cubic metre n / 10²⁷ m⁻³POLARIZATION P = σ_b / μC m⁻²
13.18756.602585625000001
19.093759.5596678125
2512.51675
30.9062515.473832187500001
36.812518.430914375

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

Show a hint

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

Reveal the worked solution
  1. At nd = 18.385 10²⁷ m⁻³, substitute into Y_out = 0.16689*nd*ef: Y₀ = 9.20482 μC m⁻².
  2. At nd = 21.6925 10²⁷ m⁻³, the same fixed-parameter model gives Y₁ = 10.8608 μC m⁻².
  3. Signed change: ΔY_out = Y₁ − Y₀ = 1.65597 μC m⁻²; the input change is Δnd = 3.3075 10²⁷ m⁻³.
  4. Average sensitivity is ΔY_out/Δnd = 0.50067 μC m⁻² per 10²⁷ m⁻³. 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 = 9.20482 μC m⁻², Y₁ = 10.8608 μC m⁻²; ΔY_out = 1.65597 μC m⁻²; average sensitivity = 0.50067 μC m⁻² per 10²⁷ m⁻³.

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

Conductors, insulators, and polarization: infer an input with uncertainty

A conductor carries charge that is free to travel through the whole sample, an insulator holds every charge bound to its own molecule, and polarization is the small separation of bound charge that a field produces inside neutral matter. Use the supplied topic model Y_out = 0.16689*nd*ef, where Y_out is POLARIZATION P = σ_b (μC m⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nd is Molecules per cubic metre n, measured in 10²⁷ m⁻³. Keep ef (Applied field E) = 3 MV m⁻¹ fixed. For this exercise use only 13.1875 10²⁷ m⁻³ ≤ nd ≤ 36.8125 10²⁷ m⁻³ 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 = 13.817866 μC m⁻², with a bounded reading uncertainty ±0.69196 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 nd by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0023625 10²⁷ m⁻³, 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

POLARIZATION P = σ_b against nd

Predictions from the supplied Conductors, insulators, and polarization 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): POLARIZATION P = σ_b / μC m⁻²

POLARIZATION P = σ_b against ndPredictions from the supplied Conductors, insulators, and polarization 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: nd: Molecules per cubic metre n / 10²⁷ m⁻³. Vertical axis: POLARIZATION P = σ_b / μC m⁻². Exact values are available in the data table below.6.1299.32312.5215.7118.912.2418.622531.3837.76xy

Horizontal axis (x): nd: Molecules per cubic metre n / 10²⁷ m⁻³

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

Reading 1nd: Molecules per cubic metre n / 10²⁷ m⁻³: 13.19POLARIZATION P = σ_b / μC m⁻²: 6.603
Read the exact data as a table
POLARIZATION P = σ_b against nd — supplied values
nd: Molecules per cubic metre n / 10²⁷ m⁻³POLARIZATION P = σ_b / μC m⁻²
13.18756.602585625000001
19.093759.5596678125
2512.51675
30.9062515.473832187500001
36.812518.430914375

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(nd) = 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(13.1875) = 6.60259 μC m⁻² and Y_out(36.8125) = 18.4309 μC m⁻². The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives nd_line = 13.1875 + (13.8179 − 6.60259)(36.8125 − 13.1875)/(18.4309 − 6.60259) = 27.5987 10²⁷ m⁻³. 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 nd = 27.5987 10²⁷ m⁻³; substituting back gives Y_out = 13.817866 μC m⁻².
  4. The interpolation error in the input is |nd_line − nd| = 0 10²⁷ m⁻³. This is within the requested numerical tolerance.
  5. Invert both limits Y_out = 13.125906 μC m⁻² and 14.509826 μC m⁻². Sort the roots even if the response decreases: 26.2167 10²⁷ m⁻³ ≤ nd ≤ 28.9808 10²⁷ m⁻³.
  6. The lower and upper input deviations are 1.38207 10²⁷ m⁻³ and 1.38207 10²⁷ m⁻³. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer nd ≈ 27.5987 10²⁷ m⁻³; compatible interval [26.2167, 28.9808] 10²⁷ m⁻³. Endpoint interpolation gives 27.5987 10²⁷ m⁻³.

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