University Physics II · unit UPII-04
Electric Potential
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.
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.
Electric potential and potential difference: infer an input with uncertainty
The electric potential at a point is the electric potential energy per unit charge that a small test charge would have there, measured in volts, where one volt is one joule per coulomb. Use the supplied topic model Y_out = -ef*xp/100, where Y_out is V(P) − V(LEFT PLATE) (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. xp is Probe position x_P, measured in cm. Keep ef (Field E from left plate to right) = 1200 V m⁻¹; xz (Position where V is set to zero) = 5 cm fixed. For this exercise use only 1.50937 cm ≤ xp ≤ 8.99062 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 = -72.87525 V, with a bounded reading uncertainty ±5.2518 V. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown xp 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.
V(P) − V(LEFT PLATE) against xp
Predictions from the supplied Electric potential and potential difference 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): V(P) − V(LEFT PLATE) / V
Horizontal axis (x): xp: Probe position x_P / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xp: Probe position x_P / cm | V(P) − V(LEFT PLATE) / V |
|---|---|
| 1.509375 | -18.1125 |
| 3.3796874999999997 | -40.55625 |
| 5.25 | -63 |
| 7.120312499999999 | -85.44374999999998 |
| 8.990625 | -107.8875 |
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(xp) = 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(1.50937) = -18.1125 V and Y_out(8.99062) = -107.888 V. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives xp_line = 1.50937 + (-72.8752 − -18.1125)(8.99062 − 1.50937)/(-107.888 − -18.1125) = 6.07294 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 xp = 6.07294 cm; substituting back gives Y_out = -72.87525 V.
- The interpolation error in the input is |xp_line − xp| = 1.77636e-15 cm. This is within the requested numerical tolerance.
- Invert both limits Y_out = -78.12705 V and -67.62345 V. Sort the roots even if the response decreases: 5.63529 cm ≤ xp ≤ 6.51059 cm.
- The lower and upper input deviations are 0.43765 cm and 0.43765 cm. 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 xp ≈ 6.07294 cm; compatible interval [5.63529, 6.51059] cm. Endpoint interpolation gives 6.07294 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Potential of a point charge: infer an input with uncertainty
The potential of a point charge is the work per unit charge needed to bring a test charge in from infinitely far away to a distance r from it: V(r) = kq/r, a single signed number carrying the sign of the source. Use the supplied topic model Y_out = 8.99*q/(r*r), where Y_out is FIELD E(r) (V m⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Probe distance r, measured in m. Keep q (Source charge q) = 2 nC fixed. For this exercise use only 0.039125 m ≤ r ≤ 0.180875 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 = 1139.8882 V m⁻¹, with a bounded reading uncertainty ±88.546 V m⁻¹. 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 0.000014175 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.
FIELD E(r) against r
Predictions from the supplied Potential of a point 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): FIELD E(r) / V m⁻¹
Horizontal axis (x): r: Probe distance r / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Probe distance r / m | FIELD E(r) / V m⁻¹ |
|---|---|
| 0.03912500000000001 | 11745.756310669698 |
| 0.0745625 | 3234.065156553772 |
| 0.11 | 1485.9504132231407 |
| 0.1454375 | 850.0351528154848 |
| 0.180875 | 549.5821252081732 |
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(0.039125) = 11745.8 V m⁻¹ and Y_out(0.180875) = 549.582 V m⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 0.039125 + (1139.89 − 11745.8)(0.180875 − 0.039125)/(549.582 − 11745.8) = 0.173401 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 r = 0.125593 m; substituting back gives Y_out = 1139.8882 V m⁻¹.
- The interpolation error in the input is |r_line − r| = 0.0478089 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1051.3422 V m⁻¹ and 1228.4342 V m⁻¹. Sort the roots even if the response decreases: 0.120981 m ≤ r ≤ 0.130774 m.
- The lower and upper input deviations are 0.00461102 m and 0.00518192 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 r ≈ 0.125593 m; compatible interval [0.120981, 0.130774] m. Endpoint interpolation gives 0.173401 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Potential superposition and electric dipoles: infer an input with uncertainty
The potential at a point due to several charges is the algebraic sum of the potentials each charge would produce there alone — signed numbers added, with no components and no angles. Use the supplied topic model Y_out = 899*(q1/(30+0.5*d) + q2/max(30-0.5*d,1)), where Y_out is V AT s = +30 cm (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. d is Separation d, measured in cm. Keep q1 (Left charge q₁) = 8 nC; q2 (Right charge q₂) = -8 nC fixed. For this exercise use only 11.4 cm ≤ d ≤ 36.6 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 = -267.12017 V, with a bounded reading uncertainty ±25.891 V. 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 0.00252 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.
V AT s = +30 cm against d
Predictions from the supplied Potential superposition and electric dipoles 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): V AT s = +30 cm / V
Horizontal axis (x): d: Separation d / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Separation d / cm | V AT s = +30 cm / V |
|---|---|
| 11.4 | -94.5104955562472 |
| 17.7 | -154.925016201612 |
| 24 | -228.3174603174603 |
| 30.299999999999997 | -325.01851292548963 |
| 36.599999999999994 | -465.7981631894674 |
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(11.4) = -94.5105 V and Y_out(36.6) = -465.798 V. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 11.4 + (-267.12 − -94.5105)(36.6 − 11.4)/(-465.798 − -94.5105) = 23.1153 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 = 26.772 cm; substituting back gives Y_out = -267.12017 V.
- The interpolation error in the input is |d_line − d| = 3.65665 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -293.01117 V and -241.22917 V. Sort the roots even if the response decreases: 24.9622 cm ≤ d ≤ 28.433 cm.
- The lower and upper input deviations are 1.80981 cm and 1.66098 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 ≈ 26.772 cm; compatible interval [24.9622, 28.433] cm. Endpoint interpolation gives 23.1153 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.