Dielectrics and Gauss' Law β€” Question 5

Section 25-7 Β· Physics pages 4140–4141

Question 5 A homogeneous spherical dielectric has bound surface charge but no bound bulk charge in its gap Use D = Ο΅0E + P, βˆ‡Β· D = ρf, ρb = βˆ’βˆ‡Β· P and Οƒb = P Β· Λ†n, where n points out of the dielectric. A conducting sphere of radius a carries +Q and a concentric shell of radius b carries -Q. The gap is filled by a uniform linear dielectric of relative permittivity kappa. a b Λ†r Schematic; labels, not drawing lengths, specify the data. (a) Find the polarization field and bound volume density in the gap. (b) Find the bound surface densities at a and b and their total charges.
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Question 5 - Solution A homogeneous spherical dielectric has bound surface charge but no bound bulk charge in its gap a b Λ†r (a) Method and calculation Dr = Q 4Ο€r2 , Pr = (1 βˆ’1/ΞΊ) Q 4Ο€r2 . ρb = βˆ’1 r2 d dr(r2Pr) = 0 (a < r < b). (b) Interpretation and further calculation Οƒb(a) = βˆ’(1 βˆ’1/ΞΊ) Q 4Ο€a2 , Οƒb(b) = +(1 βˆ’1/ΞΊ) Q 4Ο€b2 . Qb,a = βˆ’(1 βˆ’1/ΞΊ)Q, Qb,b = +(1 βˆ’1/ΞΊ)Q, Qb,net = 0. Physical check The outward normal of the dielectric points inward at its inner boundary; missing that sign reverses the inner bound charge.

Question 5

A homogeneous spherical dielectric has bound surface charge but no bound bulk charge in its gap

Use 𝑫=Ο΅0𝑬+𝑷\mathbf D=\epsilon_0\mathbf E+\mathbf P, βˆ‡β‹…π‘«=ρf\nabla\cdot\mathbf D=\rho_f, ρb=βˆ’βˆ‡β‹…π‘·\rho_b=-\nabla\cdot\mathbf P and Οƒb=𝑷⋅𝒏̂\sigma_b=\mathbf P\cdot\hat{\mathbf n}, where n points out of the dielectric. A conducting sphere of radius a carries +Q and a concentric shell of radius b carries -Q. The gap is filled by a uniform linear dielectric of relative permittivity kappa.

Diagram shown in the original SVG worksheet.

  1. Find the polarization field and bound volume density in the gap.

  2. Find the bound surface densities at a and b and their total charges.

Question 5 - Solution

A homogeneous spherical dielectric has bound surface charge but no bound bulk charge in its gap

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Dr=Q4Ο€r2,Pr=(1βˆ’1/ΞΊ)Q4Ο€r2.D_r=\frac{Q}{4\pi r^2},\quad P_r=(1-1/\kappa)\frac{Q}{4\pi r^2}.ρb=βˆ’1r2ddr(r2Pr)=0(a<r<b).\rho_b=-\frac1{r^2}\frac d{dr}(r^2P_r)=0\quad(a<r<b).

(b) Interpretation and further calculation

Οƒb(a)=βˆ’(1βˆ’1/ΞΊ)Q4Ο€a2,Οƒb(b)=+(1βˆ’1/ΞΊ)Q4Ο€b2.\sigma_b(a)=-(1-1/\kappa)\frac{Q}{4\pi a^2},\quad\sigma_b(b)=+(1-1/\kappa)\frac{Q}{4\pi b^2}.Qb,a=βˆ’(1βˆ’1/ΞΊ)Q,Qb,b=+(1βˆ’1/ΞΊ)Q,Qb,net=0.Q_{b,a}=-(1-1/\kappa)Q,\quad Q_{b,b}=+(1-1/\kappa)Q,\quad Q_{b,\mathrm{net}}=0.

Physical check

The outward normal of the dielectric points inward at its inner boundary; missing that sign reverses the inner bound charge.