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Physics / Chapter 22: Electric Fields / Section 22-3

The Electric Field Due to a Point Charge — Question 10

Section 22-3 · Physics pages 3530–3531

Question 10 A local radial gradient provides an independent distance estimate Use k = 8.99 × 109 N m2/C2; treat the specified charges as points in vacuum. Along an outward ray from a single positive point source, a sensor measures E = 180 N/C and dE/dr = −1200 N/(C m) at P. No background field is present. r E Axes and labels specify the model; the sketch is schematic. (a) Infer the source distance without first finding its charge. (b) Find the charge and predict the leading fractional field change for a 1.00 mm outward displacement.
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Question 10 - Solution A local radial gradient provides an independent distance estimate r E (a) Method and calculation E = kQ/r2, dE dr = −2kQ/r3 = −2E/r. r = − 2E dE/dr = .300 m, Q = Er2/k ≈1.80200 nC. (b) Interpretation and further calculation ∆E E ≃1 E dE dr ∆r = −2(.00100) .300 = −.00666667. E(r + ∆r) ≃178.8 N/C. The gradient measures spatial variation and adds information beyond a single local magnitude. Physical check The inference is model-dependent: a uniform background changes E but not its radial derivative.

Question 10

A local radial gradient provides an independent distance estimate

Use k=8.99×109Nm2/C2k=8.99\times10^9\,\mathrm{N\,m^2/C^2}; treat the specified charges as points in vacuum. Along an outward ray from a single positive point source, a sensor measures E=180E=180 N/C and dE/dr=−1200dE/dr=-1200 N/(C m) at P. No background field is present.

Diagram shown in the original SVG worksheet.

  1. Infer the source distance without first finding its charge.

  2. Find the charge and predict the leading fractional field change for a 1.00 mm outward displacement.

Question 10 - Solution

A local radial gradient provides an independent distance estimate

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E=kQ/r2,dEdr=−2kQ/r3=−2E/r.E=kQ/r^2,\quad\frac{dE}{dr}=-2kQ/r^3=-2E/r.r=−2EdE/dr=.300m,Q=Er2/k≈1.80200nC.r=-\frac{2E}{dE/dr}=.300\,\mathrm m,\quad Q=Er^2/k\approx1.80200\,\mathrm{nC}.

(b) Interpretation and further calculation

ΔEE≃1EdEdrΔr=−2(.00100).300=−.00666667.\frac{\Delta E}{E}\simeq\frac1E\frac{dE}{dr}\Delta r=-\frac{2(.00100)}{.300}=-.00666667.E(r+Δr)≃178.8N/C.E(r+\Delta r)\simeq178.8\,\mathrm{N/C}.The gradient measures spatial variation and adds information beyond a single local magnitude.

Physical check

The inference is model-dependent: a uniform background changes E but not its radial derivative.

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