Question

I am reading Griffiths's Introduction to Electrodynamics (3rd edition). In the part of dieletric polarization (p.166), the author first adopts the potential for a single dipole: V(r)=14πϵ0ˆr⋅pr2,V(r)=14πϵ0r^⋅pr2,(4.8) and then uses some vector analysis transformation to get the forms of surface bound charge and volume bound charge. But later I find that Feynman in hislectureshows that one can get the formulas for surface and volume bound charges using only simple physical pictures: σpol=P⋅nσpol=P⋅n(10.12)rhopol=−∇⋅P.rhopol=−∇⋅P.(10.16) Feynman's derivation is simple and clear and he didn't refer to the form of the dipole potential. Therefore, it is clear that the form of the potential for a dipole in Griffiths's derivation is unnecessary. But how THAT form of dipole potential comes into play in the derivation of(10.12)(10.12)and(10.16)(10.16)is unclear. Can someone explain it? The link comes from the charge density distribution to generate the dipole field. To make the link explicit, I’ll use distributions. You don’t need the potential of a dipole to deduce the charge density. Recall that and electric monopole of chargeqqat positionr0r0has charge distribution:rho(r)=qδ(r−r0)rho(r)=qδ(r−r0)An ideal electric dipole of momentppat positionr0r0can be seen as the limit of two opposing infinite charges at the same location that are scaled to keep a finite dipole moment. Its charge density is therefore:rho(r)=p⋅∇r0δ(r−r0)rho(r)=p⋅∇r0δ(r−r0)This is the true definition of a dipole from which you derive Griffiths’ and Feynman’s approach. Solving the Poisson equation (settingϵ0=1ϵ0=1):ΔV+rho=0ΔV+rho=0gives you Griffiths’ formula of the dipole field. The simplest way to see this is to applyp⋅∇r0p⋅∇r0on the monopole field (Coulomb potential) since it commutes with the Laplacian. When you haveNNdipoles indexed byii, the charge density is obtained by adding each contribution:rho(r)=∑ipi⋅∇riδ(r−ri)In the continuous limit, this becomes:rho(r)=∫P(r′)⋅∇r′δ(r−r′)d3r′=−∫∇r′⋅P(r′)δ(r−r′)d3r′=−∇⋅P(r) You can actually deduce (10.6) from (10.12). Indeed, a discontinuity isPwill give a Dirac delta supported on the surface of discontinuity which amounts to surface charges given by (10.6). Hope this helps.

A) Feynman's approach outlines the physical principles of surface and volume bound charges without any reliance on dipole potential. Griffiths's formulation of dipole potential is unrelated to deriving surface and volume bound charges.

B) The derivation of surface and volume bound charges, as presented by Feynman, directly incorporates Griffiths's dipole potential formulation, illustrating its significance in establishing the relationship between charge distributions and the dipole field.

C) Both Griffiths's and Feynman's approaches independently arrive at the formulas for surface and volume bound charges without any connection to the form of the dipole potential used in Griffiths's derivation.

D) The derivation of surface and volume bound charges in Feynman's approach implicitly utilizes the dipole potential specified by Griffiths to elucidate the relationship between charge distributions and the resulting dipole field.

Answer 1

Feynman's approach outlines the physical principles of surface and volume bound charges without any reliance on the dipole potential formulated by Griffiths. Griffiths's dipole potential is unrelated to the derivation of surface and volume bound charges in Feynman's presentation.Thus, the correct option is A.

Step-by-step explanation:

Firstly, Feynman's approach to deriving surface and volume bound charges is grounded in physical principles without direct reference to the specific form of the dipole potential used by Griffiths. Feynman's methodology relies on simple physical pictures and fundamental concepts such as the polarization vector, making it independent of Griffiths's dipole potential formulation.

Secondly, Griffiths, in his derivation, starts with the potential for a single dipole,
`\(V(r) = (1)/(4\pi\epsilon_0)\frac{\hat{r}\cdot\mathbf{p}}{r^2}\)`. This dipole potential is used to link the charge density distribution to generate the dipole field. However, Feynman's derivation directly incorporates physical insights without relying on the dipole potential, making his approach distinct from Griffiths's formulation.

In conclusion, option A is the correct choice as it accurately reflects that Feynman's approach to surface and volume bound charges is independent of Griffiths's dipole potential. Feynman's clear and simple physical explanation provides an alternative perspective that does not rely on the specific mathematical formulation used by Griffiths.