The Electric Field Due to a Point Charge — Question 7

Section 22-3 · Physics pages 3524–3525

Question 7 A nonradial residual rejects the hypothesis of a single source at the origin Use k = 8.99 × 109 N m2/C2; treat the specified charges as points in vacuum. At P= (.300, .400, 0) m, a measured field is (6, 8, 2) N/C. A proposed model places one point charge at the origin and no other sources. x y P Schematic; labels, not drawing lengths, specify the data. (a) Test the direction predicted by that model and find the charge that reproduces the radial projection of the measurement. (b) Find the residual field not explained by that charge and its fraction of the measured magnitude.
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Question 7 - Solution A nonradial residual rejects the hypothesis of a single source at the origin x y P (a) Method and calculation r = .500 m, ˆr = (.6, .8, 0), Er = E · ˆr = 10.0 N/C. Q = Err2/k ≈2.78087 × 10−10 C, Emodel = (6, 8, 0) N/C. (b) Interpretation and further calculation Eresidual = (0, 0, 2) N/C, |Eresidual| |E| = 2 √ 104 ≈.196116. Emeasured̸ ∥r. A charge adjustment can change radial magnitude but cannot create the required out-of-plane component. Physical check The residual identifies failure of the stipulated one-source model, not the location of any omitted source.

Question 7

A nonradial residual rejects the hypothesis of a single source at the origin

Use k=8.99×109Nm2/C2k=8.99\times10^9\,\mathrm{N\,m^2/C^2}; treat the specified charges as points in vacuum. At P=(.300,.400,0)=(.300,.400,0) m, a measured field is (6,8,2)(6,8,2) N/C. A proposed model places one point charge at the origin and no other sources.

Diagram shown in the original SVG worksheet.

  1. Test the direction predicted by that model and find the charge that reproduces the radial projection of the measurement.

  2. Find the residual field not explained by that charge and its fraction of the measured magnitude.

Question 7 - Solution

A nonradial residual rejects the hypothesis of a single source at the origin

Diagram shown in the original SVG worksheet.

(a) Method and calculation

r=.500m,𝒓̂=(.6,.8,0),Er=𝑬⋅𝒓̂=10.0N/C.r=.500\,\mathrm m,\quad\hat{\mathbf r}=(.6,.8,0),\quad E_r=\mathbf E\cdot\hat{\mathbf r}=10.0\,\mathrm{N/C}.Q=Err2/k≈2.78087×10−10C,𝑬model=(6,8,0)N/C.Q=E_rr^2/k\approx2.78087\times10^{-10}\,\mathrm C,\quad\mathbf E_{\mathrm{model}}=(6,8,0)\,\mathrm{N/C}.

(b) Interpretation and further calculation

𝑬residual=(0,0,2)N/C,|𝑬residual||𝑬|=2104≈.196116.\mathbf E_{\mathrm{residual}}=(0,0,2)\,\mathrm{N/C},\quad\frac{|\mathbf E_{\mathrm{residual}}|}{|\mathbf E|}=\frac2{\sqrt{104}}\approx.196116.𝑬measured∦𝒓.\mathbf E_{\mathrm{measured}}\not\parallel\mathbf r.A charge adjustment can change radial magnitude but cannot create the required out-of-plane component.

Physical check

The residual identifies failure of the stipulated one-source model, not the location of any omitted source.