University Physics II · topic UPII-01.05
Force superposition and charge equilibrium
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.
Force superposition and equilibrium: predict a finite change
The principle of superposition says the net electrostatic force on a charge is the vector sum of the Coulomb forces each other charge would exert on it alone, with no charge screening or altering any other pair's contribution. Use the supplied topic model Y_out = qb/qa, where Y_out is CHARGE RATIO q₂/q₁ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. qa is Left charge q₁, measured in nC. Keep qb (Right charge q₂) = 2 nC fixed. For this exercise use only 1.85 nC ≤ qa ≤ 8.15 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 qa = 3.236 nC and 4.118 nC. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δqa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CHARGE RATIO q₂/q₁ against qa
Predictions from the supplied Force superposition and equilibrium 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 RATIO q₂/q₁
Horizontal axis (x): qa: Left charge q₁ / nC
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| qa: Left charge q₁ / nC | CHARGE RATIO q₂/q₁ |
|---|---|
| 1.85 | 1.081081081081081 |
| 3.425 | 0.5839416058394161 |
| 5 | 0.4 |
| 6.574999999999999 | 0.30418250950570347 |
| 8.149999999999999 | 0.24539877300613502 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of qa separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At qa = 3.236 nC, substitute into Y_out = qb/qa: Y₀ = 0.618047.
- At qa = 4.118 nC, the same fixed-parameter model gives Y₁ = 0.485673.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.132374; the input change is Δqa = 0.882 nC.
- Average sensitivity is ΔY_out/Δqa = -0.150084 1 per nC. 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 = 0.618047, Y₁ = 0.485673; ΔY_out = -0.132374; average sensitivity = -0.150084 1 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.
Force superposition and equilibrium: infer an input with uncertainty
The principle of superposition says the net electrostatic force on a charge is the vector sum of the Coulomb forces each other charge would exert on it alone, with no charge screening or altering any other pair's contribution. Use the supplied topic model Y_out = qb/qa, where Y_out is CHARGE RATIO q₂/q₁ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. qa is Left charge q₁, measured in nC. Keep qb (Right charge q₂) = 2 nC fixed. For this exercise use only 1.85 nC ≤ qa ≤ 8.15 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 = 0.35130862, with a bounded reading uncertainty ±0.015886. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown qa 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 RATIO q₂/q₁ against qa
Predictions from the supplied Force superposition and equilibrium 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 RATIO q₂/q₁
Horizontal axis (x): qa: Left charge q₁ / nC
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| qa: Left charge q₁ / nC | CHARGE RATIO q₂/q₁ |
|---|---|
| 1.85 | 1.081081081081081 |
| 3.425 | 0.5839416058394161 |
| 5 | 0.4 |
| 6.574999999999999 | 0.30418250950570347 |
| 8.149999999999999 | 0.24539877300613502 |
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(qa) = 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(1.85) = 1.08108 and Y_out(8.15) = 0.245399. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives qa_line = 1.85 + (0.351309 − 1.08108)(8.15 − 1.85)/(0.245399 − 1.08108) = 7.35157 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 qa = 5.693 nC; substituting back gives Y_out = 0.35130862.
- The interpolation error in the input is |qa_line − qa| = 1.65857 nC. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.33542262 and 0.36719462. Sort the roots even if the response decreases: 5.4467 nC ≤ qa ≤ 5.96263 nC.
- The lower and upper input deviations are 0.246297 nC and 0.269627 nC. 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 qa ≈ 5.693 nC; compatible interval [5.4467, 5.96263] nC. Endpoint interpolation gives 7.35157 nC.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.