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Physics / Chapter 22: Electric Fields / Section 22-6

The Electric Field Due to a Charged Disk — Question 5

Section 22-6 · Physics pages 3580–3581

Question 5 A central hole removes the surface-limit field on the axis A uniformly charged annulus has inner radius a and outer radius R, with 0 < a < R, and density σ > 0. Regard it as a radius-R disk minus a radius-a disk with the same density. R a Schematic; labels, not drawing lengths, specify the data. (a) Derive its positive-axis field. (b) Find the limits z →0 at fixed a and a →0 at fixed positive z. Explain why the order of taking both limits matters.
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Question 5 - Solution A central hole removes the surface-limit field on the axis R a (a) Method and calculation E(z) = 2πkσ [( 1 − z √ z2 + R2 ) − ( 1 − z √ z2 + a2 )] . E(z) = 2πkσz ( 1 √ z2 + a2 − 1 √ z2 + R2 ) . (b) Interpretation and further calculation lim z→0+ E(z) = 0 (a > 0), lim a→0 E(z) = Edisk(z). lim a→0 lim z→0+ E = 0, lim z→0+ lim a→0 E = 2πkσ. Closing the hole introduces charged material arbitrarily near the observation point, so these limits need not commute. Physical check The annulus center is in an uncharged hole, whereas the disk center lies on an ideal charged surface.

Question 5

A central hole removes the surface-limit field on the axis

A uniformly charged annulus has inner radius a and outer radius R, with 0<a<R0<a<R, and density σ>0\sigma>0. Regard it as a radius-R disk minus a radius-a disk with the same density.

Diagram shown in the original SVG worksheet.

  1. Derive its positive-axis field.

  2. Find the limits z→0z\to0 at fixed a and a→0a\to0 at fixed positive z. Explain why the order of taking both limits matters.

Question 5 - Solution

A central hole removes the surface-limit field on the axis

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E(z)=2πkσ[(1−zz2+R2)−(1−zz2+a2)].E(z)=2\pi k\sigma\left[\left(1-\frac z{\sqrt{z^2+R^2}}\right)-\left(1-\frac z{\sqrt{z^2+a^2}}\right)\right].E(z)=2πkσz(1z2+a2−1z2+R2).E(z)=2\pi k\sigma z\left(\frac1{\sqrt{z^2+a^2}}-\frac1{\sqrt{z^2+R^2}}\right).

(b) Interpretation and further calculation

limz→0+E(z)=0(a>0),lima→0E(z)=Edisk(z).\lim_{z\to0^+}E(z)=0\quad(a>0),\qquad\lim_{a\to0}E(z)=E_{\mathrm{disk}}(z).lima→0limz→0+E=0,limz→0+lima→0E=2πkσ.\lim_{a\to0}\lim_{z\to0^+}E=0,\quad\lim_{z\to0^+}\lim_{a\to0}E=2\pi k\sigma.Closing the hole introduces charged material arbitrarily near the observation point, so these limits need not commute.

Physical check

The annulus center is in an uncharged hole, whereas the disk center lies on an ideal charged surface.

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