University Physics II · topic UPII-01.08
Electrostatic applications and safety
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.
Electrostatic Applications & Safety: predict a finite change
Electrostatic safety is the practice of keeping the largest field anywhere in an arrangement below the value at which the surrounding gas breaks down, and keeping the energy any unavoidable spark releases below what the surroundings can tolerate. Use the supplied topic model Y_out = vkv/max(rtip,0.1), where Y_out is FIELD AT TIP (MV m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. rtip is Tip radius r, measured in mm. Keep rdome (Dome radius R) = 0.2 m; vkv (Conductor potential V) = 100 kV fixed. For this exercise use only 2.57187 mm ≤ rtip ≤ 17.9281 mm 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 rtip = 5.95025 mm and 8.10012 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrtip across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FIELD AT TIP against rtip
Predictions from the supplied Electrostatic Applications & Safety 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): FIELD AT TIP / MV m⁻¹
Horizontal axis (x): rtip: Tip radius r / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rtip: Tip radius r / mm | FIELD AT TIP / MV m⁻¹ |
|---|---|
| 2.571875 | 38.88213851761847 |
| 6.410937499999999 | 15.598342676090667 |
| 10.249999999999998 | 9.756097560975611 |
| 14.089062499999999 | 7.097704336253743 |
| 17.928124999999998 | 5.577828133170647 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rtip separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rtip = 5.95025 mm, substitute into Y_out = vkv/max(rtip,0.1): Y₀ = 16.806 MV m⁻¹.
- At rtip = 8.10012 mm, the same fixed-parameter model gives Y₁ = 12.3455 MV m⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -4.46053 MV m⁻¹; the input change is Δrtip = 2.14987 mm.
- Average sensitivity is ΔY_out/Δrtip = -2.07478 MV m⁻¹ per mm. 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 = 16.806 MV m⁻¹, Y₁ = 12.3455 MV m⁻¹; ΔY_out = -4.46053 MV m⁻¹; average sensitivity = -2.07478 MV m⁻¹ per mm.
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.
Electrostatic Applications & Safety: infer an input with uncertainty
Electrostatic safety is the practice of keeping the largest field anywhere in an arrangement below the value at which the surrounding gas breaks down, and keeping the energy any unavoidable spark releases below what the surroundings can tolerate. Use the supplied topic model Y_out = vkv/max(rtip,0.1), where Y_out is FIELD AT TIP (MV m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. rtip is Tip radius r, measured in mm. Keep rdome (Dome radius R) = 0.2 m; vkv (Conductor potential V) = 100 kV fixed. For this exercise use only 2.57187 mm ≤ rtip ≤ 17.9281 mm 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 = 8.3757793 MV m⁻¹, with a bounded reading uncertainty ±0.41969 MV 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 rtip by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00153562 mm, 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.
FIELD AT TIP against rtip
Predictions from the supplied Electrostatic Applications & Safety 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): FIELD AT TIP / MV m⁻¹
Horizontal axis (x): rtip: Tip radius r / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rtip: Tip radius r / mm | FIELD AT TIP / MV m⁻¹ |
|---|---|
| 2.571875 | 38.88213851761847 |
| 6.410937499999999 | 15.598342676090667 |
| 10.249999999999998 | 9.756097560975611 |
| 14.089062499999999 | 7.097704336253743 |
| 17.928124999999998 | 5.577828133170647 |
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(rtip) = 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(2.57187) = 38.8821 MV m⁻¹ and Y_out(17.9281) = 5.57783 MV m⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rtip_line = 2.57187 + (8.37578 − 38.8821)(17.9281 − 2.57187)/(5.57783 − 38.8821) = 16.638 mm. 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 rtip = 11.9392 mm; substituting back gives Y_out = 8.3757793 MV m⁻¹.
- The interpolation error in the input is |rtip_line − rtip| = 4.69883 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 7.9560893 MV m⁻¹ and 8.7954693 MV m⁻¹. Sort the roots even if the response decreases: 11.3695 mm ≤ rtip ≤ 12.569 mm.
- The lower and upper input deviations are 0.569698 mm and 0.629802 mm. 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 rtip ≈ 11.9392 mm; compatible interval [11.3695, 12.569] mm. Endpoint interpolation gives 16.638 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.