University Physics II · topic UPII-04.01
Electric potential energy
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.
Electric potential energy: predict a finite change
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. Calculate Y_out at r = 4.15525 cm and 5.20262 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = 4.15525 cm, substitute into Y_out = 0.899*q2/(r*r): Y₀ = 0.624808 mN.
- At r = 5.20262 cm, the same fixed-parameter model gives Y₁ = 0.398562 mN.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.226246 mN; the input change is Δr = 1.04737 cm.
- Average sensitivity is ΔY_out/Δr = -0.216012 mN per cm. 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 = 0.624808 mN, Y₁ = 0.398562 mN; ΔY_out = -0.226246 mN; average sensitivity = -0.216012 mN per cm.
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.
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.