Applying Gauss' Law: Cylindrical Symmetry — Question 1

Section 23-6 · Physics pages 3732–3733

Question 1 The arbitrary Gaussian length cancels for an infinite line source Use ϵ0 = 8.854 × 10−12 C2/(N m2). An infinite straight line has uniform positive charge per length λ = 3.00 nC/m. There are no other sources. Choose a coaxial Gaussian cylinder of radius r and length L. L Finite segment of an infinitely long cylindrical model; end effects are excluded. (a) Derive the field, explicitly treating side and end fluxes. (b) Calculate the field at r = .200 m. Compare the effects of doubling the Gaussian length and doubling the observation radius.
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Question 1 - Solution The arbitrary Gaussian length cancels for an infinite line source L (a) Method and calculation Φends = 0, Φside = Er(2πrL), Qenc = λL. Er = λ 2πϵ0r. (b) Interpretation and further calculation Er(.200) = 3.00 × 10−9 2π(8.854 × 10−12)(.200) ≈269.632 N/C. Er(r; 2L) = Er(r; L), Er(2r) = 1 2Er(r). L is an arbitrary surface-construction parameter, whereas r specifies the physical observation point. Physical check Radial direction and constant side-field magnitude follow from rotational and translational symmetry of the infinite source.

Question 1

The arbitrary Gaussian length cancels for an infinite line source

Use ϵ0=8.854×10−12C2/(Nm2)\epsilon_0=8.854\times10^{-12}\,\mathrm{C^2/(N\,m^2)}. An infinite straight line has uniform positive charge per length λ=3.00\lambda=3.00 nC/m. There are no other sources. Choose a coaxial Gaussian cylinder of radius r and length L.

Diagram shown in the original SVG worksheet.

  1. Derive the field, explicitly treating side and end fluxes.

  2. Calculate the field at r=.200r=.200 m. Compare the effects of doubling the Gaussian length and doubling the observation radius.

Question 1 - Solution

The arbitrary Gaussian length cancels for an infinite line source

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Φends=0,Φside=Er(2πrL),Qenc=λL.\Phi_{\mathrm{ends}}=0,\quad\Phi_{\mathrm{side}}=E_r(2\pi rL),\quad Q_{\mathrm{enc}}=\lambda L.Er=λ2πϵ0r.E_r=\frac{\lambda}{2\pi\epsilon_0r}.

(b) Interpretation and further calculation

Er(.200)=3.00×10−92π(8.854×10−12)(.200)≈269.632N/C.E_r(.200)=\frac{3.00\times10^{-9}}{2\pi(8.854\times10^{-12})(.200)}\approx269.632\,\mathrm{N/C}.Er(r;2L)=Er(r;L),Er(2r)=12Er(r).E_r(r;2L)=E_r(r;L),\quad E_r(2r)=\tfrac12E_r(r).L is an arbitrary surface-construction parameter, whereas r specifies the physical observation point.

Physical check

Radial direction and constant side-field magnitude follow from rotational and translational symmetry of the infinite source.