Electromagnetic Field Theory: A Problem Solving Approach Part 18. Electromagnetic field theory is often the least popular course in the electrical engineering curriculum. Heavy reliance on vector and integral calculus can obscure physical phenomena so that the student becomes bogged down in the mathematics and loses sight of the applications. This book instills problem solving confidence by teaching through the use of a large number of worked problems. To keep the subject exciting, many of these problems are based on physical processes, devices, and models. This text is an introductory treatment on the junior level for a two-semester electrical engineering. | Polarization 145 b The Local Electric Field If this dipole were isolated the local electric field would equal the applied macroscopic field. However a large number density N of neighboring dipoles also contributes to the polarizing electric field. The electric field changes drastically from point to point within a small volume containing many dipoles being equal to the superposition of fields due to each dipole given by 5 . The macroscopic field is then the average field over this small volume. We calculate this average field by first finding the average field due to a single point charge Q a distance a along the z axis from the center of a spherical volume with radius R much larger than the radius of the electron cloud R Ro as in Figure 3-56. The average field due to this charge over the spherical volume is 1 i I f2ir Q Hr aiz r2 sin OdrdOdd Jr o -L o -V o 4iTEo a2 r2-2ra cos fl 3 2 28 where we used the relationships 2 2 2 rQP q r 2 ra cos fl TQP nr aiz 29 Using 23 in 28 again results in the x and y components being zero when integrated over Only the z component is now nonzero Q 2tt fR r3 cos fl a r sin fl dr dO sirR3 4ire0 Je o Jr o a2 r2 2ra cos fl s 2 30 We introduce the change of variable from fl to u u r2 a2 2ar cos fl du 2ar sin 0d0 31 so that 30 can be integrated over u and r. Performing the u integration first we have r- 3Q R 8 PS I OITxC 0 f r a 2 r r2-a2-u J r_o 4a2 u3 2 dr du - 3 2 fR J_ _ 2 r2-a2 u I r o 8ttJ 3 o Jr O 4a2 U1 2 L r-a 2 - pp 2 i r2 1 - r a dr 32 SirR3eoa2 Jr o r a We were careful to be sure to take the positive square root in the lower limit of u. Then for r a the integral is zero so 146 Polarization and Conduction that the integral limits over r range from 0 to a Et 2 i 2r2 dr 8irJc Oa Jr o 4m oK 33 To form a dipole we add a negative charge -Q a small distance d below the original charge. The average electric field due to the dipole is then the superposition of 33 for both charges Q Qd E ----- a- a-d ---- ----------- 4ireoRS .
