University Physics II · topic UPII-02.03
Field superposition and null points
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.
Field superposition and null points: predict a finite change
Superposition says the electric field at a point is the vector sum of the fields each source charge would produce there on its own, and a null point is a place where that sum comes to zero. Use the supplied topic model Y_out = 8.99*(q1/(0.4/(1+s*sqrt(q2/q1)))+s*q2/abs((0.4/(1+s*sqrt(q2/q1)))-0.4)), where Y_out is POTENTIAL V THERE (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|. q2 is Size of q₂ at x = 0.40 m, measured in nC. Keep q1 (Charge q₁ at x = 0) = 8 nC; s (Sign of q₂: −1 opposite, +1 same) = -1 fixed. For this exercise use only 1.2125 nC ≤ q2 ≤ 2.7875 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 q2 = 1.559 nC and 1.7795 nC. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δq2 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
POTENTIAL V THERE against q2
Predictions from the supplied Field superposition and null points 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): POTENTIAL V THERE / V
Horizontal axis (x): q2: Size of q₂ at x = 0.40 m / nC
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q2: Size of q₂ at x = 0.40 m / nC | POTENTIAL V THERE / V |
|---|---|
| 1.2125 | 67.05495809344706 |
| 1.6062499999999997 | 54.768668272875075 |
| 1.9999999999999998 | 44.95000000000001 |
| 2.39375 | 36.895126431500834 |
| 2.7874999999999996 | 30.182235722945645 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of q2 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At q2 = 1.559 nC, substitute into Y_out = 8.99*(q1/(0.4/(1+s*sqrt(q2/q1)))+s*q2/abs((0.4/(1+s*sqrt(q2/q1)))-0.4)): Y₀ = 56.0944 V.
- At q2 = 1.7795 nC, the same fixed-parameter model gives Y₁ = 50.1951 V.
- Signed change: ΔY_out = Y₁ − Y₀ = -5.89926 V; the input change is Δq2 = 0.2205 nC.
- Average sensitivity is ΔY_out/Δq2 = -26.754 V 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 = 56.0944 V, Y₁ = 50.1951 V; ΔY_out = -5.89926 V; average sensitivity = -26.754 V 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.
Field superposition and null points: infer an input with uncertainty
Superposition says the electric field at a point is the vector sum of the fields each source charge would produce there on its own, and a null point is a place where that sum comes to zero. Use the supplied topic model Y_out = 8.99*(q1/(0.4/(1+s*sqrt(q2/q1)))+s*q2/abs((0.4/(1+s*sqrt(q2/q1)))-0.4)), where Y_out is POTENTIAL V THERE (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|. q2 is Size of q₂ at x = 0.40 m, measured in nC. Keep q1 (Charge q₁ at x = 0) = 8 nC; s (Sign of q₂: −1 opposite, +1 same) = -1 fixed. For this exercise use only 1.2125 nC ≤ q2 ≤ 2.7875 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 = 41.217924 V, with a bounded reading uncertainty ±1.6554 V. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 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.
POTENTIAL V THERE against q2
Predictions from the supplied Field superposition and null points 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): POTENTIAL V THERE / V
Horizontal axis (x): q2: Size of q₂ at x = 0.40 m / nC
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q2: Size of q₂ at x = 0.40 m / nC | POTENTIAL V THERE / V |
|---|---|
| 1.2125 | 67.05495809344706 |
| 1.6062499999999997 | 54.768668272875075 |
| 1.9999999999999998 | 44.95000000000001 |
| 2.39375 | 36.895126431500834 |
| 2.7874999999999996 | 30.182235722945645 |
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(q2) = 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(1.2125) = 67.055 V and Y_out(2.7875) = 30.1822 V. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives q2_line = 1.2125 + (41.2179 − 67.055)(2.7875 − 1.2125)/(30.1822 − 67.055) = 2.31612 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 q2 = 2.17325 nC; substituting back gives Y_out = 41.217924 V.
- The interpolation error in the input is |q2_line − q2| = 0.142866 nC. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 39.562524 V and 42.873324 V. Sort the roots even if the response decreases: 2.09458 nC ≤ q2 ≤ 2.25501 nC.
- The lower and upper input deviations are 0.0786655 nC and 0.0817562 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 q2 ≈ 2.17325 nC; compatible interval [2.09458, 2.25501] nC. Endpoint interpolation gives 2.31612 nC.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.