A Charged Isolated Conductor — Question 8

Section 23-5 · Physics pages 3726–3727

Question 8 Nested conducting shells transmit an enclosed-charge constraint through the gaps A charge q is centered within two concentric conducting shells. Their radial intervals are a < r < b and c < r < d, with a < b < c < d, and net charges are QA and QB. There are no external sources. q a b c d Schematic; labels, not drawing lengths, specify the data. (a) Find the charges on all four conductor surfaces. (b) Find the field in b < r < c and r > d. Determine the choice of QB that makes the exterior field vanish, and state whether that also forces the gap field to vanish.
Show solutionHide solution
Question 8 - Solution Nested conducting shells transmit an enclosed-charge constraint through the gaps q a b c d (a) Method and calculation Qa = −q, Qb = QA + q. Qc = −(q + QA), Qd = QB + q + QA. (b) Interpretation and further calculation Er = q + QA 4πϵ0r2 (b < r < c), Er = q + QA + QB 4πϵ0r2 (r > d). QB = −(q + QA) ⇒Eexterior = 0. The gap field remains nonzero unless QA = −q independently. Physical check Each metal region has zero field, but each vacuum gap responds to the total charge enclosed within that gap.

Question 8

Nested conducting shells transmit an enclosed-charge constraint through the gaps

A charge q is centered within two concentric conducting shells. Their radial intervals are a<r<ba<r<b and c<r<dc<r<d, with a<b<c<da<b<c<d, and net charges are QAQ_A and QBQ_B. There are no external sources.

Diagram shown in the original SVG worksheet.

  1. Find the charges on all four conductor surfaces.

  2. Find the field in b<r<cb<r<c and r>dr>d. Determine the choice of QBQ_B that makes the exterior field vanish, and state whether that also forces the gap field to vanish.

Question 8 - Solution

Nested conducting shells transmit an enclosed-charge constraint through the gaps

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Qa=−q,Qb=QA+q.Q_a=-q,\quad Q_b=Q_A+q.Qc=−(q+QA),Qd=QB+q+QA.Q_c=-(q+Q_A),\quad Q_d=Q_B+q+Q_A.

(b) Interpretation and further calculation

Er=q+QA4πϵ0r2(b<r<c),Er=q+QA+QB4πϵ0r2(r>d).E_r=\frac{q+Q_A}{4\pi\epsilon_0r^2}\quad(b<r<c),\quad E_r=\frac{q+Q_A+Q_B}{4\pi\epsilon_0r^2}\quad(r>d).QB=−(q+QA)⇒Eexterior=0.Q_B=-(q+Q_A)\Rightarrow E_{\mathrm{exterior}}=0.The gap field remains nonzero unless QA=−qQ_A=-q independently.

Physical check

Each metal region has zero field, but each vacuum gap responds to the total charge enclosed within that gap.