University Physics II · topic UPII-01.01
Charge, conservation, and quantisation
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.
Charge, conservation and quantisation: predict a finite change
Electric charge is the signed, conserved property of matter that comes only in whole multiples of the elementary charge e = 1.602176634 × 10⁻¹⁹ C, so any object's net charge is q = N e for some integer N — positive, negative or zero. Use the supplied topic model Y_out = qa0+1.602*n, where Y_out is CHARGE ON A (nC). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. qa0 is Object A starting charge, measured in nC. Keep qb0 (Object B starting charge) = 0 nC; n (Electrons moved A → B) = 1.5 ×10¹⁰ fixed. For this exercise use only -3.15 nC ≤ qa0 ≤ 3.15 nC 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 qa0 = -1.764 nC and -0.882 nC. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δqa0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CHARGE ON A against qa0
Predictions from the supplied Charge, conservation and quantisation 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): CHARGE ON A / nC
Horizontal axis (x): qa0: Object A starting charge / nC
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| qa0: Object A starting charge / nC | CHARGE ON A / nC |
|---|---|
| -3.15 | -0.7469999999999999 |
| -1.5750000000000002 | 0.8279999999999998 |
| -4.440892098500626e-16 | 2.4029999999999996 |
| 1.5749999999999997 | 3.9779999999999998 |
| 3.149999999999999 | 5.552999999999999 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of qa0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At qa0 = -1.764 nC, substitute into Y_out = qa0+1.602*n: Y₀ = 0.639 nC.
- At qa0 = -0.882 nC, the same fixed-parameter model gives Y₁ = 1.521 nC.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.882 nC; the input change is Δqa0 = 0.882 nC.
- Average sensitivity is ΔY_out/Δqa0 = 1 nC per nC. Its positive sign means that this response increases 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 = 0.639 nC, Y₁ = 1.521 nC; ΔY_out = 0.882 nC; average sensitivity = 1 nC per nC.
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.
Charge, conservation and quantisation: infer an input with uncertainty
Electric charge is the signed, conserved property of matter that comes only in whole multiples of the elementary charge e = 1.602176634 × 10⁻¹⁹ C, so any object's net charge is q = N e for some integer N — positive, negative or zero. Use the supplied topic model Y_out = qa0+1.602*n, where Y_out is CHARGE ON A (nC). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. qa0 is Object A starting charge, measured in nC. Keep qb0 (Object B starting charge) = 0 nC; n (Electrons moved A → B) = 1.5 ×10¹⁰ fixed. For this exercise use only -3.15 nC ≤ qa0 ≤ 3.15 nC 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 = 3.096 nC, with a bounded reading uncertainty ±0.36855 nC. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown qa0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 nC, 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.
CHARGE ON A against qa0
Predictions from the supplied Charge, conservation and quantisation 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): CHARGE ON A / nC
Horizontal axis (x): qa0: Object A starting charge / nC
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| qa0: Object A starting charge / nC | CHARGE ON A / nC |
|---|---|
| -3.15 | -0.7469999999999999 |
| -1.5750000000000002 | 0.8279999999999998 |
| -4.440892098500626e-16 | 2.4029999999999996 |
| 1.5749999999999997 | 3.9779999999999998 |
| 3.149999999999999 | 5.552999999999999 |
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(qa0) = 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(-3.15) = -0.747 nC and Y_out(3.15) = 5.553 nC. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives qa0_line = -3.15 + (3.096 − -0.747)(3.15 − -3.15)/(5.553 − -0.747) = 0.693 nC. 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 qa0 = 0.693 nC; substituting back gives Y_out = 3.096 nC.
- The interpolation error in the input is |qa0_line − qa0| = 2.22045e-16 nC. This is within the requested numerical tolerance.
- Invert both limits Y_out = 2.72745 nC and 3.46455 nC. Sort the roots even if the response decreases: 0.32445 nC ≤ qa0 ≤ 1.06155 nC.
- The lower and upper input deviations are 0.36855 nC and 0.36855 nC. 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 qa0 ≈ 0.693 nC; compatible interval [0.32445, 1.06155] nC. Endpoint interpolation gives 0.693 nC.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.