A Charged Isolated Conductor — Question 1

Section 23-5 · Physics pages 3712–3713

Question 1 A surface-field limit constrains the charge on an isolated sphere An isolated conducting sphere of radius R carries charge Q and is in electrostatic equilibrium with no external sources. A specified model permits a maximum surface field magnitude Elim. R conductor Cross-section of concentric spherical surfaces; labels specify radii. (a) Derive its uniform surface density and the field immediately outside using Gauss law. (b) Find the largest allowed |Q|. Compare the changes in surface field and charge capacity when R doubles.
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Question 1 - Solution A surface-field limit constrains the charge on an isolated sphere R conductor (a) Method and calculation σ = Q 4πR2 , Er(R+)4πR2 = Q/ϵ0. Er(R+) = Q 4πϵ0R2 = σ ϵ0 , E(R−) = 0. (b) Interpretation and further calculation |Q|max = 4πϵ0R2Elim. Es(2R; Q) = Es(R; Q)/4, |Q|max(2R) = 4|Q|max(R). The threshold is a supplied field limit, not a universal breakdown value. Physical check The exterior field has a discontinuity at the charged surface; the zero interior value must not be used as the exterior field.

Question 1

A surface-field limit constrains the charge on an isolated sphere

An isolated conducting sphere of radius R carries charge Q and is in electrostatic equilibrium with no external sources. A specified model permits a maximum surface field magnitude ElimE_{\mathrm{lim}}.

Diagram shown in the original SVG worksheet.

  1. Derive its uniform surface density and the field immediately outside using Gauss law.

  2. Find the largest allowed |Q||Q|. Compare the changes in surface field and charge capacity when R doubles.

Question 1 - Solution

A surface-field limit constrains the charge on an isolated sphere

Diagram shown in the original SVG worksheet.

(a) Method and calculation

σ=Q4πR2,Er(R+)4πR2=Q/ϵ0.\sigma=\frac Q{4\pi R^2},\quad E_r(R^+)4\pi R^2=Q/\epsilon_0.Er(R+)=Q4πϵ0R2=σϵ0,𝑬(R−)=𝟎.E_r(R^+)=\frac Q{4\pi\epsilon_0R^2}=\frac\sigma{\epsilon_0},\quad\mathbf E(R^-)=\mathbf0.

(b) Interpretation and further calculation

|Q|max=4πϵ0R2Elim.|Q|_{\max}=4\pi\epsilon_0R^2E_{\mathrm{lim}}.Es(2R;Q)=Es(R;Q)/4,|Q|max(2R)=4|Q|max(R).E_s(2R;Q)=E_s(R;Q)/4,\quad |Q|_{\max}(2R)=4|Q|_{\max}(R).The threshold is a supplied field limit, not a universal breakdown value.

Physical check

The exterior field has a discontinuity at the charged surface; the zero interior value must not be used as the exterior field.