University Physics II · topic UPII-05.08
Dielectric strength and capacitor design
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.
Dielectric Strength & Capacitor Design: predict a finite change
Dielectric strength is the largest electric field an insulator can hold without conducting. Multiplied by the plate separation it gives the breakdown voltage, and it caps the energy density any design using that material can reach. Use the supplied topic model Y_out = 23.3746/d^2, where Y_out is CAPACITANCE C (μF). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Film thickness d, measured in μm. Keep em (Dielectric strength E_max) = 600 V μm⁻¹ fixed. For this exercise use only 3.95625 μm ≤ d ≤ 11.0437 μm 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 = 5.5155 μm and 6.50775 μm. 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.
CAPACITANCE C against d
Predictions from the supplied Dielectric Strength & Capacitor Design 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): CAPACITANCE C / μF
Horizontal axis (x): d: Film thickness d / μm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Film thickness d / μm | CAPACITANCE C / μF |
|---|---|
| 3.95625 | 1.4934020150291127 |
| 5.7281249999999995 | 0.7123922962931217 |
| 7.5 | 0.41554844444444444 |
| 9.271875 | 0.2718999024092566 |
| 11.04375 | 0.19165098426186689 |
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 = 5.5155 μm, substitute into Y_out = 23.3746/d^2: Y₀ = 0.768377 μF.
- At d = 6.50775 μm, the same fixed-parameter model gives Y₁ = 0.551928 μF.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.216449 μF; the input change is Δd = 0.99225 μm.
- Average sensitivity is ΔY_out/Δd = -0.21814 μF per μm. 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 = 0.768377 μF, Y₁ = 0.551928 μF; ΔY_out = -0.216449 μF; average sensitivity = -0.21814 μF per μm.
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.
Dielectric Strength & Capacitor Design: infer an input with uncertainty
Dielectric strength is the largest electric field an insulator can hold without conducting. Multiplied by the plate separation it gives the breakdown voltage, and it caps the energy density any design using that material can reach. Use the supplied topic model Y_out = 23.3746/d^2, where Y_out is CAPACITANCE C (μF). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Film thickness d, measured in μm. Keep em (Dielectric strength E_max) = 600 V μm⁻¹ fixed. For this exercise use only 3.95625 μm ≤ d ≤ 11.0437 μm 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 = 0.34097525 μF, with a bounded reading uncertainty ±0.022399 μF. 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 0.00070875 μm, 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.
CAPACITANCE C against d
Predictions from the supplied Dielectric Strength & Capacitor Design 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): CAPACITANCE C / μF
Horizontal axis (x): d: Film thickness d / μm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Film thickness d / μm | CAPACITANCE C / μF |
|---|---|
| 3.95625 | 1.4934020150291127 |
| 5.7281249999999995 | 0.7123922962931217 |
| 7.5 | 0.41554844444444444 |
| 9.271875 | 0.2718999024092566 |
| 11.04375 | 0.19165098426186689 |
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(3.95625) = 1.4934 μF and Y_out(11.0437) = 0.191651 μF. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 3.95625 + (0.340975 − 1.4934)(11.0437 − 3.95625)/(0.191651 − 1.4934) = 10.2307 μm. 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 = 8.27962 μm; substituting back gives Y_out = 0.34097525 μF.
- The interpolation error in the input is |d_line − d| = 1.95112 μm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.31857625 μF and 0.36337425 μF. Sort the roots even if the response decreases: 8.02038 μm ≤ d ≤ 8.56575 μm.
- The lower and upper input deviations are 0.259244 μm and 0.286125 μm. 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 ≈ 8.27962 μm; compatible interval [8.02038, 8.56575] μm. Endpoint interpolation gives 10.2307 μm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.