Applying Gauss' Law: Cylindrical Symmetry — Question 10

Section 23-6 · Physics pages 3750–3751

Question 10 Circular motion around an infinite line has a radius-independent speed A particle of mass m and charge −q with q > 0 moves in the transverse plane of an infinite line with λ > 0. Ignore radiation, gravity and other forces. Define C = qλ/(2πϵ0) > 0, so the inward force magnitude is C/r. r0 +λ v Circular orbit in a plane perpendicular to the charged line. (a) Find the speed and period of a circular orbit of radius r0. (b) For fixed angular momentum ℓ, use m¨r = ℓ2/(mr3) −C/r to find the frequency of small radial oscillations about the circular orbit and assess radial stability.
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Question 10 - Solution Circular motion around an infinite line has a radius-independent speed r0 +λ v (a) Method and calculation mv2 r0 = C r0 ⇒v = q C/m = s qλ 2πϵ0m. T = 2πr0 q m/C, Ω= v/r0. (b) Interpretation and further calculation ℓ2 = mCr2 0, d dr ℓ2 mr3 −C r ! r0 = −2C r2 0 . r = r0 + ξ : ¨ξ + 2C mr2 0 ξ = 0, ωr = √ 2 Ω. The restoring linear coefficient makes the orbit radially stable within the stated classical model. Physical check The line field falls as 1/r, so the r factors cancel in the circular speed but not in the orbital period.

Question 10

Circular motion around an infinite line has a radius-independent speed

A particle of mass m and charge −q-q with q>0q>0 moves in the transverse plane of an infinite line with λ>0\lambda>0. Ignore radiation, gravity and other forces. Define C=qλ/(2πϵ0)>0C=q\lambda/(2\pi\epsilon_0)>0, so the inward force magnitude is C/r.

Diagram shown in the original SVG worksheet.

  1. Find the speed and period of a circular orbit of radius r0r_0.

  2. For fixed angular momentum ℓ\ell, use mr̈=ℓ2/(mr3)−C/rm\ddot r=\ell^2/(mr^3)-C/r to find the frequency of small radial oscillations about the circular orbit and assess radial stability.

Question 10 - Solution

Circular motion around an infinite line has a radius-independent speed

Diagram shown in the original SVG worksheet.

(a) Method and calculation

mv2r0=Cr0⇒v=C/m=qλ2πϵ0m.\frac{mv^2}{r_0}=\frac C{r_0}\Rightarrow v=\sqrt{C/m}=\sqrt{\frac{q\lambda}{2\pi\epsilon_0m}}.T=2πr0m/C,Ω=v/r0.T=2\pi r_0\sqrt{m/C},\quad\Omega=v/r_0.

(b) Interpretation and further calculation

ℓ2=mCr02,ddr(ℓ2mr3−Cr)|r0=−2Cr02.\ell^2=mCr_0^2,\quad\left.\frac{d}{dr}\left(\frac{\ell^2}{mr^3}-\frac Cr\right)\right|_{r_0}=-\frac{2C}{r_0^2}.r=r0+ξ:ξ̈+2Cmr02ξ=0,ωr=2Ω.r=r_0+\xi:\quad\ddot\xi+\frac{2C}{mr_0^2}\xi=0,\quad\omega_r=\sqrt2\,\Omega.The restoring linear coefficient makes the orbit radially stable within the stated classical model.

Physical check

The line field falls as 1/r1/r, so the r factors cancel in the circular speed but not in the orbital period.