Physics 217, Fall 2002 25 October 2002(c) University of Rochester 125 October 2002 Physics 217, Fall 2002 1Today in Physics 217: dielectrics Finish things from last lecture Dielectrics Electric polarization and bound charge Calculation of field and potential from uniformly-polarized objectsP+-++++++++--------++++++++--------+++++-----=25 October 2002 Physics 217, Fall 2002 2Dielectric materialsSolids are generally composed of neutral atoms and molecules, some of which have built-in, permanent dipole moments and some of which are simply polarizable. For non-conducting solids, there is zero dipole moment on large scales, since the orientation of permanent dipoles is generally random. immersion in an electric field polarizes atoms and molecules, and tends to align their permanent dipole moments. this polarizationis characterized by a dipole moment per unit volume, in the same direction as the applied field.Non-conducting solids that can be polarized in this way are called dielectrics.25 October 2002 Physics 217, Fall 2002 3Electric polarization and bound chargeThe electric polarization P is a vector quantity:The potential from a lump of polarizedmatter isNow, dddττ′==∫Pp pPPEd′′rr22ˆˆVdτ⋅⋅′==∫pPVrrrr2ˆ111 =− =−  ′=———rrrrrras we saw long time ago (lecture, 18 September)VSPhysics 217, Fall 2002 25 October 2002(c) University of Rochester 225 October 2002 Physics 217, Fall 2002 4Electric polarization and bound charge (continued)(Reminder: take the Cartesian components of to be X,Y,Z, those of r to be x,y,z, those of r’ to be x’,y’,z’. ThenSo′=−rrrXXxxXX x xX XYYyyYY y yY YZZZZzzZ z zZ    ∂∂∂ ∂ ∂∂∂ ∂  −     ′′∂∂∂∂ ∂∂∂ ∂     ∂∂∂ ∂ ∂∂∂ ∂     ′== == == =−     ′′∂∂∂∂ ∂∂∂ ∂     ∂∂    ∂∂∂ ∂∂∂ −     ∂∂  ′′∂∂∂ ∂∂∂    ———r=−—r1Vdτ′′=⋅∫P —r25 October 2002 Physics 217, Fall 2002 5Electric polarization and bound charge (continued)Let’s integrate this by parts, using product rule #5:Define the surface and volume bound chargedensities:() ()ff f⋅=⋅+⋅AAA———111.Vdddda dττττ  ′′′ ′ ′′=⋅ = ⋅ − ⋅    ′′′=⋅− ⋅∫∫∫∫∫PPPPPvVVVSV—— ——rrrrr()ˆˆ outward normal of bbσρ=⋅ ==− ⋅Pn nPS—25 October 2002 Physics 217, Fall 2002 6Electric polarization and bound charge (continued)Then the potential takes on a familiar form:Bound charge is the charge displaced by the field into dipolar form. Note that for a uniform (constant) polarization, the bound volume charge density is zero:leaving just surface bound charge, like so:.bbVdadσρτ′′=−∫∫vSVrr0bρ=−⋅ =P—d-q qd-q qd-q qq3d-qNeutralizedPhysics 217, Fall 2002 25 October 2002(c) University of Rochester 325 October 2002 Physics 217, Fall 2002 7Calculation of field and potential from uniformly-polarized objectsYou’ll be please to know that you’ve already made major progress toward calculations involving uniformly-polarized objects, because you can make them by superposition of uniformly-charged objects. Consider a uniformly polarized cylinder:P+-++++++++--------++++++++--------+++++-----Same as two uniform, oppositely charged cylinders, displaced infinitesimally along the axis (see problem 2.27, HW#4).=25 October 2002 Physics 217, Fall 2002 8Calculation of field and potential from uniformly-polarized objects (continued)Or a uniformly-polarized sphere, which would work out the same as problem 2.18 (HW #3). d+-P=Uniformly-polarized sphere, radius R. Two uniformly-charged spheres, density ±ρ, displaced by d.25 October 2002 Physics 217, Fall 2002 9Calculation of field and potential from uniformly-polarized objects (continued)We found in that problem that the field in the overlap region (or, if you like, inside the polarized sphere) isOutside, the field is just that of a simple dipole, since both spheres act like point charges located at their centers:(z axis is along P). Note that the (in principle infinitesimal) displacement distance d drops out of the problem when everything is expressed in terms of polarization.3344.33qRRπρ π=− =− =− =−dpPEd()()3334ˆˆˆˆ2cos sin 2cos sinpRPrrπθθ θθ=+= +Erθ r