The Electric Field — Question 3

Section 22-1 · Physics pages 3476–3477

Question 3 Rotating the coordinate axes changes components but not the field An electric field has components (Ex, Ey) = (60, 80) N/C. New axes are rotated counterclockwise by ϕ = 30◦, with ˆx′ = cos ϕ ˆx + sin ϕ ˆy. x y x′ y′ Schematic; labels, not drawing lengths, specify the data. (a) Find the field components in the rotated axes by projection. (b) Check the magnitude and express the force on a −2.00 nC probe in the rotated basis.
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Question 3 - Solution Rotating the coordinate axes changes components but not the field x y x′ y′ (a) Method and calculation Ex′ = Ex cos ϕ + Ey sin ϕ = 30 √ 3 + 40 ≈91.9615 N/C. Ey′ = −Ex sin ϕ + Ey cos ϕ = −30 + 40 √ 3 ≈39.2820 N/C. (b) Interpretation and further calculation (Ex′)2 + (Ey′)2 = 602 + 802 = 104 (N/C)2. F′ = (−183.923ˆx′ −78.5641ˆy′) nN. The vector has not physically rotated; only its numerical description has. Physical check An orthonormal rotation preserves dot products and therefore the squared magnitude.

Question 3

Rotating the coordinate axes changes components but not the field

An electric field has components (Ex,Ey)=(60,80)(E_x,E_y)=(60,80) N/C. New axes are rotated counterclockwise by ϕ=30∘\phi=30^\circ, with 𝒙̂′=cos⁡ϕ𝒙̂+sin⁡ϕ𝒚̂\hat{\mathbf x}\prime=\cos\phi\,\hat{\mathbf x}+\sin\phi\,\hat{\mathbf y}.

Diagram shown in the original SVG worksheet.

  1. Find the field components in the rotated axes by projection.

  2. Check the magnitude and express the force on a −2.00-2.00 nC probe in the rotated basis.

Question 3 - Solution

Rotating the coordinate axes changes components but not the field

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Ex′=Excos⁡ϕ+Eysin⁡ϕ=303+40≈91.9615N/C.E_x\prime=E_x\cos\phi+E_y\sin\phi=30\sqrt3+40\approx91.9615\,\mathrm{N/C}.Ey′=−Exsin⁡ϕ+Eycos⁡ϕ=−30+403≈39.2820N/C.E_y\prime=-E_x\sin\phi+E_y\cos\phi=-30+40\sqrt3\approx39.2820\,\mathrm{N/C}.

(b) Interpretation and further calculation

(Ex′)2+(Ey′)2=602+802=104(N/C)2.(E_x\prime)^2+(E_y\prime)^2=60^2+80^2=10^4\,\mathrm{(N/C)^2}.𝑭′=(−183.923𝒙̂′−78.5641𝒚̂′)nN.\mathbf F\prime=(-183.923\hat{\mathbf x}\prime-78.5641\hat{\mathbf y}\prime)\,\mathrm{nN}.The vector has not physically rotated; only its numerical description has.

Physical check

An orthonormal rotation preserves dot products and therefore the squared magnitude.