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11/4/2004 The Polarization Vector.doc 1/3 The Polarization Vector Recall that in dielectric materials (i.e., insulators), the charges are bound. - p + E (r ) As a result, atoms/molecules form electric dipoles when an electric field is present! Note that even for some small volume ∆v , there are many atoms/molecules present; therefore there will be many electric dipoles. E (r ) Jim Stiles The Univ. of Kansas Dept. of EECS 11/4/2004 The Polarization Vector.doc 2/3 We will therefore define an average dipole moment, per unit volume, called the Polarization Vector P ( r ) . P (r ) ∑ pn ∆v ⎡ dipole moment C ⎤ ⎢ unit volume = m 2 ⎥ ⎣ ⎦ where pn is one of N dipole moments in volume ∆v , centered at position r . Note the polarization vector is a vector field. As a result, the direction and magnitude of the Polarization vector can change as function of position (i.e., a function of r ). Q: How are vector fields P ( r ) and E ( r ) related?? A: Recall that the direction of each dipole moment is the same as the polarizing electric field. Thus P ( r ) and E ( r ) have the same direction. There magnitudes are related by a unitless scalar value χe ( r ) , called electric susceptibility: P ( r ) = ε 0 χe (r ) E (r ) Electric susceptibility is a material parameter indicating the “stretchability” of the dipoles. Jim Stiles The Univ. of Kansas Dept. of EECS 11/4/2004 The Polarization Vector.doc 3/3 Q: Can we determine the fields created by a polarized material? A: Recall the electric potential field created by one dipole is: V (r ) = p ⋅ ( r-r′ ) 4πε 0 r-r′ 3 Therefore, using d p = P ( r ) dv , the electric potential field created by a distribution of dipoles (i.e., P ( r ) ) across some volume V is (see fig. 5.9): V (r ) = ∫∫∫ V P ( r′ ) ⋅ ( r-r′ ) 4πε 0 r-r′ 3 dv ′ Q: But I thought scalar charge distributions ρv ( r ) and ρs ( r ) created the electric potential field V ( r ) . Now you are saying that electric fields are created by the vector field P ( r ) !?! A: As we will soon see, the polarization vector P ( r ) creates equivalent charge distributions—we will get the correct answer for V ( r ) from either source! Jim Stiles The Univ. of Kansas Dept. of EECS