A sphere of radius is uniformly polarised with ,

PHYSICS · Subjective · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Conductors, capacitance and boundary-value electrostatics · Electric fields in matter

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

Problem

A sphere of radius a=3.0 cma = 3.0\,\mathrm{cm} is uniformly polarised with P=Pz^\mathbf{P} = P\hat{\mathbf{z}}, P=2.0×10−6 C m−2P = 2.0\times10^{-6}\,\mathrm{C\,m^{-2}}. There is no free charge anywhere. (a) Starting from the fact that the bound surface charge is σb=Pcos⁡θ\sigma_b = P\cos\theta and the bound volume charge vanishes, argue that the interior and exterior potentials must each contain only the ℓ=1\ell = 1 Legendre term, so that Vin=Arcos⁡θV_{\rm in} = Ar\cos\theta and Vout=Bcos⁡θ/r2V_{\rm out} = B\cos\theta/r^2. (b) Impose continuity of VV at r=ar=a and the interface condition ε0(∂V/∂n)out−ε0(∂V/∂n)in=−σb\varepsilon_0(\partial V/\partial n)_{\rm out} - \varepsilon_0(\partial V/\partial n)_{\rm in} = -\sigma_b (with n^\hat{\mathbf{n}} pointing outward from the sphere) to find AA and BB. (c) Give the numerical value of Ein\mathbf{E}_{\rm in}. (d) Show that BB corresponds to a point dipole of moment p=43πa3Pp = \tfrac43\pi a^3 P sitting at the centre, give its numerical value, and check that Ein\mathbf{E}_{\rm in} times the sphere's volume equals −p/3ε0-p/3\varepsilon_0.

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