The Electric Field — Question 7

Section 22-1 · Physics pages 3484–3485

Question 7 A vanishing mean field can coexist with a nonzero force fluctuation A prescribed field switches between +E0ˆx and −E0ˆx with equal time fractions, where E0 = 200 N/C. A fixed nonperturbing probe has charge +3.00 nC. Ignore switching transients and average over many complete switches. t E0 −E0 Schematic; labels, not drawing lengths, specify the data. (a) Find the time-averaged field, its root-mean-square magnitude and the mean probe force. (b) Find the root-mean-square force. Explain why applying the mean field does not reproduce the instantaneous mechanical forcing.
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Question 7 - Solution A vanishing mean field can coexist with a nonzero force fluctuation t E0 −E0 (a) Method and calculation ⟨E⟩= 1 2(E0 −E0)ˆx = 0. Erms = √ ⟨E2x⟩= E0 = 200 N/C, ⟨F⟩= q⟨E⟩= 0. (b) Interpretation and further calculation Frms = |q|Erms = 6.00 × 10−7 N. ⟨E2 x⟩= 40000 (N/C)2̸ = ⟨Ex⟩2. A zero first moment does not remove fluctuations; motion can depend on the switching time scale. Physical check This is a time-dependent prescribed-field model, not a static field-line picture.

Question 7

A vanishing mean field can coexist with a nonzero force fluctuation

A prescribed field switches between +E0𝒙̂+E_0\hat{\mathbf x} and −E0𝒙̂-E_0\hat{\mathbf x} with equal time fractions, where E0=200E_0=200 N/C. A fixed nonperturbing probe has charge +3.00+3.00 nC. Ignore switching transients and average over many complete switches.

Diagram shown in the original SVG worksheet.

  1. Find the time-averaged field, its root-mean-square magnitude and the mean probe force.

  2. Find the root-mean-square force. Explain why applying the mean field does not reproduce the instantaneous mechanical forcing.

Question 7 - Solution

A vanishing mean field can coexist with a nonzero force fluctuation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

⟨𝑬⟩=12(E0−E0)𝒙̂=𝟎.\langle\mathbf E\rangle=\tfrac12(E_0-E_0)\hat{\mathbf x}=\mathbf0.Erms=⟨Ex2⟩=E0=200N/C,⟨𝑭⟩=q⟨𝑬⟩=𝟎.E_{\mathrm{rms}}=\sqrt{\langle E_x^2\rangle}=E_0=200\,\mathrm{N/C},\quad\langle\mathbf F\rangle=q\langle\mathbf E\rangle=\mathbf0.

(b) Interpretation and further calculation

Frms=|q|Erms=6.00×10−7N.F_{\mathrm{rms}}=|q|E_{\mathrm{rms}}=6.00\times10^{-7}\,\mathrm N.⟨Ex2⟩=40000(N/C)2≠⟨Ex⟩2.\langle E_x^2\rangle=40000\,\mathrm{(N/C)^2}\ne\langle E_x\rangle^2.A zero first moment does not remove fluctuations; motion can depend on the switching time scale.

Physical check

This is a time-dependent prescribed-field model, not a static field-line picture.