Potential of a Charged Isolated Conductor — Question 3

Section 24-11 · Physics pages 3996–3997

Question 3 Equal surface voltages give larger charge density on the smaller sphere Two separate distant conducting spheres of radii a and b have the same positive potential V relative to infinity. There are no nearby external sources. V V a b Schematic; labels, not drawing lengths, specify the data. (a) Find each charge and surface density. (b) Find their charging energies and the ratios of charge, density and energy for b=3a.
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Question 3 - Solution Equal surface voltages give larger charge density on the smaller sphere V V a b (a) Method and calculation Qa = V a/k, Qb = V b/k. σa = V 4πka, σb = V 4πkb. (b) Interpretation and further calculation Ui = ∫Qi 0 kq′ Ri dq′ = V 2Ri 2k . b = 3a : Qb/Qa = 3, σb/σa = 1/3, Ub/Ua = 3. Physical check At fixed voltage, larger radius stores more charge and energy while reducing surface density and field.

Question 3

Equal surface voltages give larger charge density on the smaller sphere

Two separate distant conducting spheres of radii a and b have the same positive potential V relative to infinity. There are no nearby external sources.

Diagram shown in the original SVG worksheet.

  1. Find each charge and surface density.

  2. Find their charging energies and the ratios of charge, density and energy for b=3a.

Question 3 - Solution

Equal surface voltages give larger charge density on the smaller sphere

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Qa=Va/k,Qb=Vb/k.Q_a=Va/k,\quad Q_b=Vb/k.σa=V4πka,σb=V4πkb.\sigma_a=\frac{V}{4\pi ka},\quad\sigma_b=\frac{V}{4\pi kb}.

(b) Interpretation and further calculation

Ui=∫0Qikq′Ridq′=V2Ri2k.U_i=\int_0^{Q_i}\frac{kq\prime}{R_i}dq\prime=\frac{V^2R_i}{2k}.b=3a:Qb/Qa=3,σb/σa=1/3,Ub/Ua=3.b=3a:\quad Q_b/Q_a=3,\quad\sigma_b/\sigma_a=1/3,\quad U_b/U_a=3.

Physical check

At fixed voltage, larger radius stores more charge and energy while reducing surface density and field.