University Physics II · unit UPII-01
Electric Charge and Coulomb's Law
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.
Conductors, insulators, and polarization: infer an input with uncertainty
A conductor carries charge that is free to travel through the whole sample, an insulator holds every charge bound to its own molecule, and polarization is the small separation of bound charge that a field produces inside neutral matter. Use the supplied topic model Y_out = 0.16689*nd*ef, where Y_out is POLARIZATION P = σ_b (μC m⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nd is Molecules per cubic metre n, measured in 10²⁷ m⁻³. Keep ef (Applied field E) = 3 MV m⁻¹ fixed. For this exercise use only 13.1875 10²⁷ m⁻³ ≤ nd ≤ 36.8125 10²⁷ m⁻³ 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 = 13.817866 μC m⁻², with a bounded reading uncertainty ±0.69196 μC m⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown nd by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0023625 10²⁷ m⁻³, 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.
POLARIZATION P = σ_b against nd
Predictions from the supplied Conductors, insulators, and polarization 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): POLARIZATION P = σ_b / μC m⁻²
Horizontal axis (x): nd: Molecules per cubic metre n / 10²⁷ m⁻³
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| nd: Molecules per cubic metre n / 10²⁷ m⁻³ | POLARIZATION P = σ_b / μC m⁻² |
|---|---|
| 13.1875 | 6.602585625000001 |
| 19.09375 | 9.5596678125 |
| 25 | 12.51675 |
| 30.90625 | 15.473832187500001 |
| 36.8125 | 18.430914375 |
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(nd) = 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(13.1875) = 6.60259 μC m⁻² and Y_out(36.8125) = 18.4309 μC m⁻². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives nd_line = 13.1875 + (13.8179 − 6.60259)(36.8125 − 13.1875)/(18.4309 − 6.60259) = 27.5987 10²⁷ m⁻³. 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 nd = 27.5987 10²⁷ m⁻³; substituting back gives Y_out = 13.817866 μC m⁻².
- The interpolation error in the input is |nd_line − nd| = 0 10²⁷ m⁻³. This is within the requested numerical tolerance.
- Invert both limits Y_out = 13.125906 μC m⁻² and 14.509826 μC m⁻². Sort the roots even if the response decreases: 26.2167 10²⁷ m⁻³ ≤ nd ≤ 28.9808 10²⁷ m⁻³.
- The lower and upper input deviations are 1.38207 10²⁷ m⁻³ and 1.38207 10²⁷ m⁻³. 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 nd ≈ 27.5987 10²⁷ m⁻³; compatible interval [26.2167, 28.9808] 10²⁷ m⁻³. Endpoint interpolation gives 27.5987 10²⁷ m⁻³.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Charging by Friction, Contact & Induction: infer an input with uncertainty
Charging by friction, contact and induction are three ways of moving electrons between objects: rubbing transfers them at a surface, touching shares them between conductors, and induction uses a nearby charge plus a path to earth to strand a charge of the opposite sign. Use the supplied topic model Y_out = 4*q/d*clamp(stage-1,0,1), where Y_out is NET ON SPHERE (nC). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; clamp(x,a,b) confines x to [a,b]. d is Bead distance from centre, measured in cm. Keep stage (Step in the sequence) = 2; q (Bead charge magnitude) = 6 nC fixed. For this exercise use only 9.7 cm ≤ d ≤ 22.3 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 = 1.380421 nC, with a bounded reading uncertainty ±0.045628 nC. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00126 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.
NET ON SPHERE against d
Predictions from the supplied Charging by Friction, Contact & Induction 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): NET ON SPHERE / nC
Horizontal axis (x): d: Bead distance from centre / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Bead distance from centre / cm | NET ON SPHERE / nC |
|---|---|
| 9.7 | 2.4742268041237114 |
| 12.849999999999998 | 1.8677042801556423 |
| 15.999999999999998 | 1.5000000000000002 |
| 19.15 | 1.2532637075718016 |
| 22.299999999999997 | 1.0762331838565025 |
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(d) = 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(9.7) = 2.47423 nC and Y_out(22.3) = 1.07623 nC. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 9.7 + (1.38042 − 2.47423)(22.3 − 9.7)/(1.07623 − 2.47423) = 19.5584 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 d = 17.386 cm; substituting back gives Y_out = 1.380421 nC.
- The interpolation error in the input is |d_line − d| = 2.17238 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.334793 nC and 1.426049 nC. Sort the roots even if the response decreases: 16.8297 cm ≤ d ≤ 17.9803 cm.
- The lower and upper input deviations are 0.556284 cm and 0.594316 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 d ≈ 17.386 cm; compatible interval [16.8297, 17.9803] cm. Endpoint interpolation gives 19.5584 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Coulomb's Law: infer an input with uncertainty
Coulomb's law says that two point charges at rest push or pull each other along the line joining them, with a force proportional to the product of their charge magnitudes and inversely proportional to the square of their separation. Use the supplied topic model Y_out = 89.88*abs(q1*q2)/(r*r), where Y_out is FORCE ON EACH F (µN). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|. r is Separation r, measured in cm. Keep q1 (Charge q₁ at the left) = 5 nC; q2 (Charge q₂ on the right) = -3 nC fixed. For this exercise use only 4.6375 cm ≤ r ≤ 9.3625 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 = 23.842265 µN, with a bounded reading uncertainty ±1.2693 µN. 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 0.0004725 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.
FORCE ON EACH F against r
Predictions from the supplied Coulomb's Law 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): FORCE ON EACH F / µN
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 | FORCE ON EACH F / µN |
|---|---|
| 4.6375 | 62.68829781823729 |
| 5.8187500000000005 | 39.81941965547595 |
| 7 | 27.51428571428571 |
| 8.18125 | 20.14257526053688 |
| 9.3625 | 15.38050734312416 |
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(4.6375) = 62.6883 µN and Y_out(9.3625) = 15.3805 µN. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 4.6375 + (23.8423 − 62.6883)(9.3625 − 4.6375)/(15.3805 − 62.6883) = 8.51736 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 = 7.51975 cm; substituting back gives Y_out = 23.842265 µN.
- The interpolation error in the input is |r_line − r| = 0.997608 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 22.572965 µN and 25.111565 µN. Sort the roots even if the response decreases: 7.32724 cm ≤ r ≤ 7.72828 cm.
- The lower and upper input deviations are 0.192513 cm and 0.20853 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 ≈ 7.51975 cm; compatible interval [7.32724, 7.72828] cm. Endpoint interpolation gives 8.51736 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.