Potential of a Charged Isolated Conductor — Question 8

Section 24-11 · Physics pages 4006–4007

Question 8 Nested isolated conductors can have zero exterior potential and a nonzero internal voltage A conducting sphere of radius a carries +Q. A concentric thin conducting shell of radius b>a carries -Q. There are no other sources; set V(infinity)=0. +Q −Q a b Schematic; labels, not drawing lengths, specify the data. (a) Find V everywhere, including within the inner sphere and outside the shell. (b) Find the voltage between the conductors and the quasistatic work moving a small positive probe q from just inside b to just outside a.
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Question 8 - Solution Nested isolated conductors can have zero exterior potential and a nonzero internal voltage +Q −Q a b (a) Method and calculation V (r) =        kQ(1/a −1/b) r ≤a, kQ(1/r −1/b) a < r < b, 0 r ≥b. Va −Vb = kQ(1/a −1/b) > 0. (b) Interpretation and further calculation Wext = q(Va −Vb) = kQq(1/a −1/b). E = kQ r2 ˆr (a < r < b), E = 0 (r < a or r > b). Physical check The total charge vanishes, so the external field and infinity-referenced potential vanish; charge separation still produces an internal voltage.

Question 8

Nested isolated conductors can have zero exterior potential and a nonzero internal voltage

A conducting sphere of radius a carries +Q. A concentric thin conducting shell of radius b>a carries -Q. There are no other sources; set V(infinity)=0.

Diagram shown in the original SVG worksheet.

  1. Find V everywhere, including within the inner sphere and outside the shell.

  2. Find the voltage between the conductors and the quasistatic work moving a small positive probe q from just inside b to just outside a.

Question 8 - Solution

Nested isolated conductors can have zero exterior potential and a nonzero internal voltage

Diagram shown in the original SVG worksheet.

(a) Method and calculation

V(r)={kQ(1/a−1/b)r≤a,kQ(1/r−1/b)a<r<b,0r≥b.V(r)=\begin{cases}kQ(1/a-1/b)&r\le a,\\kQ(1/r-1/b)&a<r<b,\\0&r\ge b.\end{cases}Va−Vb=kQ(1/a−1/b)>0.V_a-V_b=kQ(1/a-1/b)>0.

(b) Interpretation and further calculation

Wext=q(Va−Vb)=kQq(1/a−1/b).W_{\mathrm{ext}}=q(V_a-V_b)=kQq(1/a-1/b).𝑬=kQr2𝒓̂(a<r<b),𝑬=0(r<a or r>b).\mathbf E=\frac{kQ}{r^2}\hat{\mathbf r}\quad(a<r<b),\quad\mathbf E=0\quad(r<a\text{ or }r>b).

Physical check

The total charge vanishes, so the external field and infinity-referenced potential vanish; charge separation still produces an internal voltage.