University Physics II · topic UPII-03.06
Planar symmetry
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.
Planar symmetry and sheets of charge: predict a finite change
Planar symmetry is the case where a charge distribution looks identical under every slide parallel to a plane and every reflection in it, which forces the field to lie along the plane's normal with a magnitude that can depend only on the perpendicular distance. Use the supplied topic model Y_out = -56.47*(s1+s2), where Y_out is Eₓ LEFT OF BOTH (N C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s1 is Left sheet density σ₁, measured in nC m⁻². Keep s2 (Right sheet density σ₂) = -5 nC m⁻²; d (Separation of the sheets) = 8 cm fixed. For this exercise use only -15.75 nC m⁻² ≤ s1 ≤ 15.75 nC m⁻² 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 s1 = -8.82 nC m⁻² and -4.41 nC m⁻². Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs1 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
Eₓ LEFT OF BOTH against s1
Predictions from the supplied Planar symmetry and sheets of charge 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): Eₓ LEFT OF BOTH / N C⁻¹
Horizontal axis (x): s1: Left sheet density σ₁ / nC m⁻²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s1: Left sheet density σ₁ / nC m⁻² | Eₓ LEFT OF BOTH / N C⁻¹ |
|---|---|
| -15.75 | 1171.7525 |
| -7.875 | 727.05125 |
| 0 | 282.35 |
| 7.875 | -162.35125 |
| 15.75 | -607.0525 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s1 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s1 = -8.82 nC m⁻², substitute into Y_out = -56.47*(s1+s2): Y₀ = 780.415 N C⁻¹.
- At s1 = -4.41 nC m⁻², the same fixed-parameter model gives Y₁ = 531.383 N C⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -249.033 N C⁻¹; the input change is Δs1 = 4.41 nC m⁻².
- Average sensitivity is ΔY_out/Δs1 = -56.47 N C⁻¹ per nC m⁻². 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 = 780.415 N C⁻¹, Y₁ = 531.383 N C⁻¹; ΔY_out = -249.033 N C⁻¹; average sensitivity = -56.47 N C⁻¹ per nC m⁻².
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.
Planar symmetry and sheets of charge: infer an input with uncertainty
Planar symmetry is the case where a charge distribution looks identical under every slide parallel to a plane and every reflection in it, which forces the field to lie along the plane's normal with a magnitude that can depend only on the perpendicular distance. Use the supplied topic model Y_out = -56.47*(s1+s2), where Y_out is Eₓ LEFT OF BOTH (N C⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s1 is Left sheet density σ₁, measured in nC m⁻². Keep s2 (Right sheet density σ₂) = -5 nC m⁻²; d (Separation of the sheets) = 8 cm fixed. For this exercise use only -15.75 nC m⁻² ≤ s1 ≤ 15.75 nC m⁻² 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 = 86.68145 N C⁻¹, with a bounded reading uncertainty ±104.06 N C⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00315 nC m⁻², 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.
Eₓ LEFT OF BOTH against s1
Predictions from the supplied Planar symmetry and sheets of charge 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): Eₓ LEFT OF BOTH / N C⁻¹
Horizontal axis (x): s1: Left sheet density σ₁ / nC m⁻²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s1: Left sheet density σ₁ / nC m⁻² | Eₓ LEFT OF BOTH / N C⁻¹ |
|---|---|
| -15.75 | 1171.7525 |
| -7.875 | 727.05125 |
| 0 | 282.35 |
| 7.875 | -162.35125 |
| 15.75 | -607.0525 |
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(s1) = 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(-15.75) = 1171.75 N C⁻¹ and Y_out(15.75) = -607.053 N C⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s1_line = -15.75 + (86.6814 − 1171.75)(15.75 − -15.75)/(-607.053 − 1171.75) = 3.465 nC m⁻². 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 s1 = 3.465 nC m⁻²; substituting back gives Y_out = 86.68145 N C⁻¹.
- The interpolation error in the input is |s1_line − s1| = 0 nC m⁻². This is within the requested numerical tolerance.
- Invert both limits Y_out = -17.37855 N C⁻¹ and 190.74145 N C⁻¹. Sort the roots even if the response decreases: 1.62225 nC m⁻² ≤ s1 ≤ 5.30775 nC m⁻².
- The lower and upper input deviations are 1.84275 nC m⁻² and 1.84275 nC m⁻². 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 s1 ≈ 3.465 nC m⁻²; compatible interval [1.62225, 5.30775] nC m⁻². Endpoint interpolation gives 3.465 nC m⁻².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.