Calculating the Capacitance — Question 10

Section 25-2 · Physics pages 4050–4051

Question 10 A coaxial geometry has an optimal inner radius for a specified field ceiling A long vacuum coaxial capacitor has fixed outer radius b and variable inner radius a. Its voltage is V>0. A prescribed maximum allowable field is E*. Ignore end effects. a b Schematic; labels, not drawing lengths, specify the data. (a) Express the greatest field in terms of V,a,b and obtain the limiting voltage. (b) Choose a to maximize that voltage for fixed b and E*, and find the corresponding capacitance per length.
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Question 10 - Solution A coaxial geometry has an optimal inner radius for a specified field ceiling a b (a) Method and calculation E(r) = V r ln(b/a), Emax = E(a). Vlim(a) = E∗a ln(b/a). (b) Interpretation and further calculation dVlim da = E∗[ln(b/a) −1] = 0 ⇒a = b/e. d2Vlim da2 = −E∗/a < 0, Vmax = E∗b/e, C/L = 2πϵ0. Physical check The largest allowable voltage does not occur at the largest capacitance; making the inner radius too large squeezes the gap.

Question 10

A coaxial geometry has an optimal inner radius for a specified field ceiling

A long vacuum coaxial capacitor has fixed outer radius b and variable inner radius a. Its voltage is V>0. A prescribed maximum allowable field is E*. Ignore end effects.

Diagram shown in the original SVG worksheet.

  1. Express the greatest field in terms of V,a,b and obtain the limiting voltage.

  2. Choose a to maximize that voltage for fixed b and E*, and find the corresponding capacitance per length.

Question 10 - Solution

A coaxial geometry has an optimal inner radius for a specified field ceiling

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E(r)=Vrln⁡(b/a),Emax=E(a).E(r)=\frac{V}{r\ln(b/a)},\quad E_{\max}=E(a).Vlim(a)=E*aln⁡(b/a).V_{\mathrm{lim}}(a)=E_*a\ln(b/a).

(b) Interpretation and further calculation

dVlimda=E*[ln⁡(b/a)−1]=0⇒a=b/e.\frac{dV_{\mathrm{lim}}}{da}=E_*[\ln(b/a)-1]=0\Rightarrow a=b/e.d2Vlimda2=−E*/a<0,Vmax=E*b/e,C/L=2πϵ0.\frac{d^2V_{\mathrm{lim}}}{da^2}=-E_*/a<0,\quad V_{\max}=E_*b/e,\quad C/L=2\pi\epsilon_0.

Physical check

The largest allowable voltage does not occur at the largest capacitance; making the inner radius too large squeezes the gap.