The Electric Field — Question 9

Section 22-1 · Physics pages 3488–3489

Question 9 Different local fields may produce the same force magnitude At P the field is (30, 40) N/C; at R it is (−40, 30) N/C. A +2.00 nC probe can be placed at either point without changing the source. A sensor records only force magnitude. x y P R Schematic; labels, not drawing lengths, specify the data. (a) Find the force magnitudes and the angle between the two field vectors. (b) Find the component-sensitive force change when moving the probe from P to R. State what additional measurement would distinguish the locations.
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Question 9 - Solution Different local fields may produce the same force magnitude x y P R (a) Method and calculation |EP | = |ER| = 50 N/C ⇒|FP | = |FR| = 100 nN. EP · ER = 30(−40) + 40(30) = 0 ⇒θ = 90◦. (b) Interpretation and further calculation ∆E = (−70, −10) N/C, ∆F = (−140, −20) nN. |∆F| = 100 √ 2 nN. One signed directional component, such as Fx, distinguishes the locations even though a magnitude-only sensor does not. Physical check The change of a vector magnitude is not generally the magnitude of the vector change.

Question 9

Different local fields may produce the same force magnitude

At P the field is (30,40)(30,40) N/C; at R it is (−40,30)(-40,30) N/C. A +2.00+2.00 nC probe can be placed at either point without changing the source. A sensor records only force magnitude.

Diagram shown in the original SVG worksheet.

  1. Find the force magnitudes and the angle between the two field vectors.

  2. Find the component-sensitive force change when moving the probe from P to R. State what additional measurement would distinguish the locations.

Question 9 - Solution

Different local fields may produce the same force magnitude

Diagram shown in the original SVG worksheet.

(a) Method and calculation

|𝑬P|=|𝑬R|=50N/C⇒|𝑭P|=|𝑭R|=100nN.|\mathbf E_P|=|\mathbf E_R|=50\,\mathrm{N/C}\Rightarrow|\mathbf F_P|=|\mathbf F_R|=100\,\mathrm{nN}.𝑬P⋅𝑬R=30(−40)+40(30)=0⇒θ=90∘.\mathbf E_P\cdot\mathbf E_R=30(-40)+40(30)=0\Rightarrow\theta=90^\circ.

(b) Interpretation and further calculation

Δ𝑬=(−70,−10)N/C,Δ𝑭=(−140,−20)nN.\Delta\mathbf E=(-70,-10)\,\mathrm{N/C},\quad\Delta\mathbf F=(-140,-20)\,\mathrm{nN}.|Δ𝑭|=1002nN.|\Delta\mathbf F|=100\sqrt2\,\mathrm{nN}.One signed directional component, such as FxF_x, distinguishes the locations even though a magnitude-only sensor does not.

Physical check

The change of a vector magnitude is not generally the magnitude of the vector change.