The Electric Field — Question 5

Section 22-1 · Physics pages 3480–3481

Question 5 Two oblique sensor projections reconstruct a planar field A nonperturbing sensor measures a planar field projection of 40.0 N/C along +x and 70.0 N/C along a direction 60◦counterclockwise from +x. x y 60◦ Schematic; labels, not drawing lengths, specify the data. (a) Reconstruct Ex, Ey and the field magnitude. (b) The second direction is changed to 0◦while the first direction stays fixed. Explain algebraically why the two readings then cannot determine Ey.
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Question 5 - Solution Two oblique sensor projections reconstruct a planar field x y 60◦ (a) Method and calculation Ex = 40.0, 70.0 = Ex cos 60◦+ Ey sin 60◦. Ey = 100/ √ 3 ≈57.7350 N/C, |E| = √ 1600 + 10000/3 ≈70.2377 N/C. (b) Interpretation and further calculation Eϕ = Ex cos ϕ + Ey sin ϕ ⇒Ey = Eϕ −Ex cos ϕ sin ϕ . ϕ = 0 : sin ϕ = 0, Eϕ = Ex. Parallel sensor directions give redundant information. Near-parallel directions also amplify measurement error in the inferred transverse component. Physical check The second reading is a projection, not the vertical component.

Question 5

Two oblique sensor projections reconstruct a planar field

A nonperturbing sensor measures a planar field projection of 40.0 N/C along +x+x and 70.0 N/C along a direction 60∘60^\circ counterclockwise from +x+x.

Diagram shown in the original SVG worksheet.

  1. Reconstruct Ex,EyE_x,E_y and the field magnitude.

  2. The second direction is changed to 0∘0^\circ while the first direction stays fixed. Explain algebraically why the two readings then cannot determine EyE_y.

Question 5 - Solution

Two oblique sensor projections reconstruct a planar field

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Ex=40.0,70.0=Excos⁡60∘+Eysin⁡60∘.E_x=40.0,\quad70.0=E_x\cos60^\circ+E_y\sin60^\circ.Ey=100/3≈57.7350N/C,|𝑬|=1600+10000/3≈70.2377N/C.E_y=100/\sqrt3\approx57.7350\,\mathrm{N/C},\quad|\mathbf E|=\sqrt{1600+10000/3}\approx70.2377\,\mathrm{N/C}.

(b) Interpretation and further calculation

Eϕ=Excos⁡ϕ+Eysin⁡ϕ⇒Ey=Eϕ−Excos⁡ϕsin⁡ϕ.E_{\phi}=E_x\cos\phi+E_y\sin\phi\Rightarrow E_y=\frac{E_{\phi}-E_x\cos\phi}{\sin\phi}.ϕ=0:sin⁡ϕ=0,Eϕ=Ex.\phi=0:\quad\sin\phi=0,\quad E_\phi=E_x.Parallel sensor directions give redundant information. Near-parallel directions also amplify measurement error in the inferred transverse component.

Physical check

The second reading is a projection, not the vertical component.