University Physics II · topic UPII-02.04
Electric field lines
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 field lines: predict a finite change
An electric field line is a curve drawn so that the electric field is tangent to it at every point, arrowed in the direction of the force on a positive test charge. Use the supplied topic model Y_out = deg(atan2(py/hypot(px,py)^3 + q*py/hypot(px-8,py)^3, px/hypot(px,py)^3 + q*(px-8)/hypot(px-8,py)^3)), where Y_out is TANGENT ANGLE AT P (° from +x). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²); deg converts radians to degrees; atan2(y,x) is the angle of the vector (x,y), in radians. py is Probe P, height above the axis, measured in cm. Keep q (Charge on B, in units of q_A) = -1; px (Probe P, distance right of A) = 2 cm fixed. For this exercise use only 0.8975 cm ≤ py ≤ 3.1025 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 py = 1.3826 cm and 1.6913 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δpy across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
TANGENT ANGLE AT P against py
Predictions from the supplied Electric field lines 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): TANGENT ANGLE AT P / ° from +x
Horizontal axis (x): py: Probe P, height above the axis / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| py: Probe P, height above the axis / cm | TANGENT ANGLE AT P / ° from +x |
|---|---|
| 0.8975 | 20.53419956537583 |
| 1.44875 | 29.62716580032207 |
| 1.9999999999999998 | 35.67529651749391 |
| 2.5512499999999996 | 39.262782912852295 |
| 3.1024999999999996 | 41.07004898158468 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of py separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At py = 1.3826 cm, substitute into Y_out = deg(atan2(py/hypot(px,py)^3 + q*py/hypot(px-8,py)^3, px/hypot(px,py)^3 + q*(px-8)/hypot(px-8,py)^3)): Y₀ = 28.7044 ° from +x.
- At py = 1.6913 cm, the same fixed-parameter model gives Y₁ = 32.6359 ° from +x.
- Signed change: ΔY_out = Y₁ − Y₀ = 3.93155 ° from +x; the input change is Δpy = 0.3087 cm.
- Average sensitivity is ΔY_out/Δpy = 12.7358 ° from +x per cm. Its positive sign means that this response increases 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 = 28.7044 ° from +x, Y₁ = 32.6359 ° from +x; ΔY_out = 3.93155 ° from +x; average sensitivity = 12.7358 ° from +x 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 field lines: infer an input with uncertainty
An electric field line is a curve drawn so that the electric field is tangent to it at every point, arrowed in the direction of the force on a positive test charge. Use the supplied topic model Y_out = deg(atan2(py/hypot(px,py)^3 + q*py/hypot(px-8,py)^3, px/hypot(px,py)^3 + q*(px-8)/hypot(px-8,py)^3)), where Y_out is TANGENT ANGLE AT P (° from +x). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²); deg converts radians to degrees; atan2(y,x) is the angle of the vector (x,y), in radians. py is Probe P, height above the axis, measured in cm. Keep q (Charge on B, in units of q_A) = -1; px (Probe P, distance right of A) = 2 cm fixed. For this exercise use only 0.8975 cm ≤ py ≤ 3.1025 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 = 37.51514 ° from +x, with a bounded reading uncertainty ±0.53324 ° from +x. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown py by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0002205 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.
TANGENT ANGLE AT P against py
Predictions from the supplied Electric field lines 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): TANGENT ANGLE AT P / ° from +x
Horizontal axis (x): py: Probe P, height above the axis / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| py: Probe P, height above the axis / cm | TANGENT ANGLE AT P / ° from +x |
|---|---|
| 0.8975 | 20.53419956537583 |
| 1.44875 | 29.62716580032207 |
| 1.9999999999999998 | 35.67529651749391 |
| 2.5512499999999996 | 39.262782912852295 |
| 3.1024999999999996 | 41.07004898158468 |
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(py) = 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(0.8975) = 20.5342 ° from +x and Y_out(3.1025) = 41.07 ° from +x. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives py_line = 0.8975 + (37.5151 − 20.5342)(3.1025 − 0.8975)/(41.07 − 20.5342) = 2.7208 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 py = 2.24255 cm; substituting back gives Y_out = 37.51514 ° from +x.
- The interpolation error in the input is |py_line − py| = 0.478248 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 36.9819 ° from +x and 38.04838 ° from +x. Sort the roots even if the response decreases: 2.16595 cm ≤ py ≤ 2.32591 cm.
- The lower and upper input deviations are 0.076601 cm and 0.0833617 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 py ≈ 2.24255 cm; compatible interval [2.16595, 2.32591] cm. Endpoint interpolation gives 2.7208 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.