Flux of an Electric Field — Question 5

Section 23-2 · Physics pages 3660–3661

Question 5 A point source sends a finite fraction of its flux through a disk A positive point charge q is at (0, 0, h) with h > 0. A disk of radius R lies in z = 0 and is oriented downward. Use Coulomb field constant k = 1/(4πϵ0). q h ˆn Schematic; labels, not drawing lengths, specify the data. (a) Integrate the electric flux through the disk, treating it as rings of radius r. (b) Find R/h for which the disk flux is πkq, and examine the small-disk and infinite-disk limits.
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Question 5 - Solution A point source sends a finite fraction of its flux through a disk q h ˆn (a) Method and calculation En(r) = kqh (r2 + h2)3/2 , dA = 2πr dr. Φ = 2πkqh ∫R 0 r dr (r2 + h2)3/2 = 2πkq ( 1 − h √ h2 + R2 ) . (b) Interpretation and further calculation Φ = πkq ⇒ h √ h2 + R2 = 1 2 ⇒R/h = √ 3. R ≪h : Φ ≃kqπR2 h2 ; R →∞: Φ →2πkq. Physical check The downward normal makes the flux positive because the charge field at the disk points away from the source, downward.

Question 5

A point source sends a finite fraction of its flux through a disk

A positive point charge q is at (0,0,h)(0,0,h) with h>0h>0. A disk of radius R lies in z=0z=0 and is oriented downward. Use Coulomb field constant k=1/(4πϵ0)k=1/(4\pi\epsilon_0).

Diagram shown in the original SVG worksheet.

  1. Integrate the electric flux through the disk, treating it as rings of radius r.

  2. Find R/h for which the disk flux is πkq\pi kq, and examine the small-disk and infinite-disk limits.

Question 5 - Solution

A point source sends a finite fraction of its flux through a disk

Diagram shown in the original SVG worksheet.

(a) Method and calculation

En(r)=kqh(r2+h2)3/2,dA=2πrdr.E_n(r)=\frac{kqh}{(r^2+h^2)^{3/2}},\quad dA=2\pi r\,dr.Φ=2πkqh∫0Rrdr(r2+h2)3/2=2πkq(1−hh2+R2).\Phi=2\pi kqh\int_0^R\frac{r\,dr}{(r^2+h^2)^{3/2}}=2\pi kq\left(1-\frac h{\sqrt{h^2+R^2}}\right).

(b) Interpretation and further calculation

Φ=πkq⇒hh2+R2=12⇒R/h=3.\Phi=\pi kq\Rightarrow\frac h{\sqrt{h^2+R^2}}=\tfrac12\Rightarrow R/h=\sqrt3.R≪h:Φ≃kqπR2h2;R→∞:Φ→2πkq.R\ll h:\ \Phi\simeq\frac{kq\pi R^2}{h^2};\qquad R\to\infty:\ \Phi\to2\pi kq.

Physical check

The downward normal makes the flux positive because the charge field at the disk points away from the source, downward.