A long thin rod-shaped electret has radius

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electric fields in matter · Electric fields in matter

The Boundary-Value Problem, Ch Polarisation and bound charge, completed

Problem

A long thin rod-shaped electret has radius a=2.0 cma = 2.0\,\mathrm{cm}, length L=50 cmL = 50\,\mathrm{cm} (so L≫aL \gg a, and the rod may be treated as a needle polarised along its own axis), and is uniformly polarised along its length with P=P0z^\mathbf{P} = P_0\hat{\mathbf{z}}, P0=2.0×10−5 C m−2P_0 = 2.0\times10^{-5}\,\mathrm{C\,m^{-2}}. There is no free charge anywhere on or in the rod. Someone works through it as follows: "The two flat ends carry bound surface charge σb=P⋅n^=±P0\sigma_b = \mathbf{P}\cdot\hat{\mathbf{n}} = \pm P_0, so charges +P0πa2+P_0\pi a^2 and −P0πa2-P_0\pi a^2. The curved side also carries bound charge, since P≠0\mathbf{P}\ne0 there: σb=P0\sigma_b = P_0 over the whole curved area 2πaL2\pi a L, giving an extra charge 2πaLP02\pi a L P_0. Adding the three contributions, the total bound charge on the rod is 2πaLP0≠02\pi a L P_0 \ne 0: bound charge is evidently not neutral for a rod, only for special shapes such as the sphere. Now, since the rod carries no free charge at all, ρf=0\rho_f = 0 and σf=0\sigma_f = 0 everywhere, so by the defining relation D=εE\mathbf{D} = \varepsilon\mathbf{E} for a linear medium, and because there is no free charge to drive E\mathbf{E}, both D\mathbf{D} and E\mathbf{E} must vanish together throughout the rod." ((a)) Find the error in the calculation of the curved-surface bound charge, and give the correct total bound charge on the rod. ((b)) Find the error in the final claim that D=εE\mathbf{D} = \varepsilon\mathbf{E} forces both fields to vanish inside the electret, and state what E\mathbf{E} and D\mathbf{D} actually are inside this long rod.

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