University Physics II · course UPII
University Physics II
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
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.
Electric field & the test charge: infer an input with uncertainty
The electric field at a point is the force a charge would feel there divided by that charge, taken in the limit of a vanishingly small positive test charge, so that the result describes the point and not the probe. Use the supplied topic model Y_out = 6+9.98617*q/k, where Y_out is SEPARATION (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Stiffness of the source mount, measured in mN m⁻¹. Keep q (Probe charge q₀) = 2 nC fixed. For this exercise use only 60.625 mN m⁻¹ ≤ k ≤ 139.375 mN m⁻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 = 6.1838016 cm, with a bounded reading uncertainty ±0.0060753 cm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.007875 mN m⁻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.
SEPARATION against k
Predictions from the supplied Electric field & the test 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): SEPARATION / cm
Horizontal axis (x): k: Stiffness of the source mount / mN m⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Stiffness of the source mount / mN m⁻¹ | SEPARATION / cm |
|---|---|
| 60.625 | 6.329440659793814 |
| 80.3125 | 6.248682832684825 |
| 100 | 6.1997234 |
| 119.6875 | 6.166870725848564 |
| 139.375 | 6.14329930044843 |
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(k) = 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(60.625) = 6.32944 cm and Y_out(139.375) = 6.1433 cm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 60.625 + (6.1838 − 6.32944)(139.375 − 60.625)/(6.1433 − 6.32944) = 122.24 mN m⁻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 k = 108.662 mN m⁻¹; substituting back gives Y_out = 6.1838016 cm.
- The interpolation error in the input is |k_line − k| = 13.5774 mN m⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 6.1777263 cm and 6.1898769 cm. Sort the roots even if the response decreases: 105.186 mN m⁻¹ ≤ k ≤ 112.377 mN m⁻¹.
- The lower and upper input deviations are 3.47676 mN m⁻¹ and 3.71446 mN m⁻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 k ≈ 108.662 mN m⁻¹; compatible interval [105.186, 112.377] mN m⁻¹. Endpoint interpolation gives 122.24 mN m⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Electric Flux: infer an input with uncertainty
The electric flux through a surface is the field component perpendicular to it, summed over the whole surface: Φ_E = ∫ E⋅dA, a signed scalar measured in N m² C⁻¹ that counts how much field crosses. Use the supplied topic model Y_out = s*s, where Y_out is PANEL AREA A (m²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Panel side, measured in m. Keep t (Angle θ from normal to field) = 55 °; ef (Field magnitude E) = 300 N C⁻¹ fixed. For this exercise use only 0.2425 m ≤ s ≤ 0.5575 m 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 = 0.18892062 m², with a bounded reading uncertainty ±0.018283 m². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000315 m, 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.
PANEL AREA A against s
Predictions from the supplied Electric Flux 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): PANEL AREA A / m²
Horizontal axis (x): s: Panel side / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Panel side / m | PANEL AREA A / m² |
|---|---|
| 0.2425 | 0.05880625 |
| 0.32125 | 0.10320156249999998 |
| 0.3999999999999999 | 0.15999999999999992 |
| 0.4787499999999999 | 0.2292015624999999 |
| 0.5574999999999999 | 0.3108062499999999 |
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(s) = 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(0.2425) = 0.0588062 m² and Y_out(0.5575) = 0.310806 m². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives s_line = 0.2425 + (0.188921 − 0.0588062)(0.5575 − 0.2425)/(0.310806 − 0.0588062) = 0.405143 m. 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 s = 0.43465 m; substituting back gives Y_out = 0.18892062 m².
- The interpolation error in the input is |s_line − s| = 0.029507 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.17063762 m² and 0.20720362 m². Sort the roots even if the response decreases: 0.413083 m ≤ s ≤ 0.455196 m.
- The lower and upper input deviations are 0.0215669 m and 0.0205462 m. 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 s ≈ 0.43465 m; compatible interval [0.413083, 0.455196] m. Endpoint interpolation gives 0.405143 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Electric potential energy: infer an input with uncertainty
The electric potential energy of a set of charges is the work an outside agent must do against the electric forces to assemble them from infinite separation, bringing each one in slowly so that none is left moving. Use the supplied topic model Y_out = 0.899*q2/(r*r), where Y_out is RADIAL FORCE F_r (mN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Separation r, measured in cm. Keep q2 (Bead charge q₂) = 12 nC; et (Total energy E) = 60 μJ fixed. For this exercise use only 2.50937 cm ≤ r ≤ 9.99062 cm 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 = 0.21564595 mN, with a bounded reading uncertainty ±0.016135 mN. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000748125 cm, 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.
RADIAL FORCE F_r against r
Predictions from the supplied Electric potential energy 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): RADIAL FORCE F_r / mN
Horizontal axis (x): r: Separation r / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Separation r / cm | RADIAL FORCE F_r / mN |
|---|---|
| 2.509375 | 1.7132068566040488 |
| 4.379687499999999 | 0.5624121446760386 |
| 6.25 | 0.2761728 |
| 8.120312499999999 | 0.16360457804711054 |
| 9.990625 | 0.10808255980519742 |
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(r) = 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(2.50937) = 1.71321 mN and Y_out(9.99062) = 0.108083 mN. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 2.50937 + (0.215646 − 1.71321)(9.99062 − 2.50937)/(0.108083 − 1.71321) = 9.48929 cm. 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 r = 7.07294 cm; substituting back gives Y_out = 0.21564595 mN.
- The interpolation error in the input is |r_line − r| = 2.41635 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.19951095 mN and 0.23178095 mN. Sort the roots even if the response decreases: 6.82231 cm ≤ r ≤ 7.35338 cm.
- The lower and upper input deviations are 0.250625 cm and 0.280444 cm. 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 r ≈ 7.07294 cm; compatible interval [6.82231, 7.35338] cm. Endpoint interpolation gives 9.48929 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.