University Physics II · topic UPII-02.07
Uniform fields and parallel plates
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.
Uniform fields and parallel plates: predict a finite change
A uniform electric field is one whose field vector is identical at every point of a region — same magnitude and same direction — so a charge placed anywhere in it feels exactly the same force. Use the supplied topic model Y_out = 0.7914*ef/(v0^2), where Y_out is EXIT DEFLECTION y (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v0 is Entry speed v₀, measured in ×10⁷ m s⁻¹. Keep ef (Field strength E) = 20 kN C⁻¹; d (Plate gap d) = 8 mm fixed. For this exercise use only 1.425 ×10⁷ m s⁻¹ ≤ v0 ≤ 4.575 ×10⁷ m s⁻¹ 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 v0 = 2.118 ×10⁷ m s⁻¹ and 2.559 ×10⁷ m s⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
EXIT DEFLECTION y against v0
Predictions from the supplied Uniform fields and parallel plates 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): EXIT DEFLECTION y / mm
Horizontal axis (x): v0: Entry speed v₀ / ×10⁷ m s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: Entry speed v₀ / ×10⁷ m s⁻¹ | EXIT DEFLECTION y / mm |
|---|---|
| 1.425 | 7.794644506001846 |
| 2.2125 | 3.2334003638801114 |
| 3 | 1.7586666666666666 |
| 3.7874999999999996 | 1.103368950756462 |
| 4.574999999999999 | 0.7562124876825229 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v0 = 2.118 ×10⁷ m s⁻¹, substitute into Y_out = 0.7914*ef/(v0^2): Y₀ = 3.52837 mm.
- At v0 = 2.559 ×10⁷ m s⁻¹, the same fixed-parameter model gives Y₁ = 2.41705 mm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.11132 mm; the input change is Δv0 = 0.441 ×10⁷ m s⁻¹.
- Average sensitivity is ΔY_out/Δv0 = -2.52 mm per ×10⁷ m s⁻¹. 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 = 3.52837 mm, Y₁ = 2.41705 mm; ΔY_out = -1.11132 mm; average sensitivity = -2.52 mm per ×10⁷ m s⁻¹.
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.
Uniform fields and parallel plates: infer an input with uncertainty
A uniform electric field is one whose field vector is identical at every point of a region — same magnitude and same direction — so a charge placed anywhere in it feels exactly the same force. Use the supplied topic model Y_out = 0.7914*ef/(v0^2), where Y_out is EXIT DEFLECTION y (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v0 is Entry speed v₀, measured in ×10⁷ m s⁻¹. Keep ef (Field strength E) = 20 kN C⁻¹; d (Plate gap d) = 8 mm fixed. For this exercise use only 1.425 ×10⁷ m s⁻¹ ≤ v0 ≤ 4.575 ×10⁷ m s⁻¹ 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 = 1.4133326 mm, with a bounded reading uncertainty ±0.098568 mm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315 ×10⁷ m s⁻¹, 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.
EXIT DEFLECTION y against v0
Predictions from the supplied Uniform fields and parallel plates 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): EXIT DEFLECTION y / mm
Horizontal axis (x): v0: Entry speed v₀ / ×10⁷ m s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: Entry speed v₀ / ×10⁷ m s⁻¹ | EXIT DEFLECTION y / mm |
|---|---|
| 1.425 | 7.794644506001846 |
| 2.2125 | 3.2334003638801114 |
| 3 | 1.7586666666666666 |
| 3.7874999999999996 | 1.103368950756462 |
| 4.574999999999999 | 0.7562124876825229 |
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(v0) = 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(1.425) = 7.79464 mm and Y_out(4.575) = 0.756212 mm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v0_line = 1.425 + (1.41333 − 7.79464)(4.575 − 1.425)/(0.756212 − 7.79464) = 4.28091 ×10⁷ m s⁻¹. 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 v0 = 3.3465 ×10⁷ m s⁻¹; substituting back gives Y_out = 1.4133326 mm.
- The interpolation error in the input is |v0_line − v0| = 0.934411 ×10⁷ m s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.3147646 mm and 1.5119006 mm. Sort the roots even if the response decreases: 3.23557 ×10⁷ m s⁻¹ ≤ v0 ≤ 3.46968 ×10⁷ m s⁻¹.
- The lower and upper input deviations are 0.110926 ×10⁷ m s⁻¹ and 0.123177 ×10⁷ m s⁻¹. 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 v0 ≈ 3.3465 ×10⁷ m s⁻¹; compatible interval [3.23557, 3.46968] ×10⁷ m s⁻¹. Endpoint interpolation gives 4.28091 ×10⁷ m s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.