A Charged Isolated Conductor — Question 3

Section 23-5 · Physics pages 3716–3717

Question 3 Separate cavities impose separate induced-charge constraints A connected isolated conductor of net charge +5.00 nC contains two closed cavities. Their isolated contents have net charges q1 = +4.00 nC and q2 = −6.00 nC. No contents touch the metal. q1 q2 Schematic; labels, not drawing lengths, specify the data. (a) Find the induced charge on each cavity wall and on the outer surface. (b) An external insulating mechanism adds +2.00 nC to the second cavity contents without transferring charge to the conductor. Find all three new surface totals and their changes.
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Question 3 - Solution Separate cavities impose separate induced-charge constraints q1 q2 (a) Method and calculation Qw1 = −q1 = −4.00 nC, Qw2 = −q2 = +6.00 nC. Qo = Qc + q1 + q2 = +3.00 nC. (b) Interpretation and further calculation q2′ = −4.00 nC ⇒Qw1′ = −4.00, Qw2′ = +4.00, Qo′ = +5.00 (nC). ∆Qw1 = 0, ∆Qw2 = −2.00, ∆Qo = +2.00 (nC). The metal redistributes charge while retaining its specified net total. Physical check The two separate Gaussian surfaces in the metal each enforce zero enclosed charge around their respective cavity.

Question 3

Separate cavities impose separate induced-charge constraints

A connected isolated conductor of net charge +5.00+5.00 nC contains two closed cavities. Their isolated contents have net charges q1=+4.00q_1=+4.00 nC and q2=−6.00q_2=-6.00 nC. No contents touch the metal.

Diagram shown in the original SVG worksheet.

  1. Find the induced charge on each cavity wall and on the outer surface.

  2. An external insulating mechanism adds +2.00+2.00 nC to the second cavity contents without transferring charge to the conductor. Find all three new surface totals and their changes.

Question 3 - Solution

Separate cavities impose separate induced-charge constraints

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Qw1=−q1=−4.00nC,Qw2=−q2=+6.00nC.Q_{w1}=-q_1=-4.00\,\mathrm{nC},\quad Q_{w2}=-q_2=+6.00\,\mathrm{nC}.Qo=Qc+q1+q2=+3.00nC.Q_o=Q_c+q_1+q_2=+3.00\,\mathrm{nC}.

(b) Interpretation and further calculation

q2′=−4.00nC⇒Qw1′=−4.00,Qw2′=+4.00,Qo′=+5.00(nC).q_2\prime=-4.00\,\mathrm{nC}\Rightarrow Q_{w1}\prime=-4.00,\quad Q_{w2}\prime=+4.00,\quad Q_o\prime=+5.00\quad(\mathrm{nC}).ΔQw1=0,ΔQw2=−2.00,ΔQo=+2.00(nC).\Delta Q_{w1}=0,\quad\Delta Q_{w2}=-2.00,\quad\Delta Q_o=+2.00\quad(\mathrm{nC}).The metal redistributes charge while retaining its specified net total.

Physical check

The two separate Gaussian surfaces in the metal each enforce zero enclosed charge around their respective cavity.