Electromagnetic FieldsThis revised edition provides patient guidance in its clear and organized presentation of problems. It is rich in variety, large in number and provides very careful treatment of relativity. One outstanding feature is the inclusion of simple, standard examples demonstrated in different methods that will allow students to enhance and understand their calculating abilities. There are over 145 worked examples; virtually all of the standard problems are included. |
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Page 73
... becomes ( 4-16 ) E , ( r ) = Q Απεργ ( r > a ) ( 4-17 ) and the electric field outside is the same as if all of the charge were a point charge located at the center of the sphere . We recall from ( 3-1 ) and ( 2-29 ) ( where the total ...
... becomes ( 4-16 ) E , ( r ) = Q Απεργ ( r > a ) ( 4-17 ) and the electric field outside is the same as if all of the charge were a point charge located at the center of the sphere . We recall from ( 3-1 ) and ( 2-29 ) ( where the total ...
Page 83
... becomes ф = ρ > = ant , So " So So - Απερ a r'2 sin 0 ' dr ' de'do ' ( z2 + r'2 — 2zr ' cos 0 ' ) 1 / 2 ( 5-17 ) where we have used ( 1-99 ) and taken the constant value of p outside of the integral . The integration over do ' can be ...
... becomes ф = ρ > = ant , So " So So - Απερ a r'2 sin 0 ' dr ' de'do ' ( z2 + r'2 — 2zr ' cos 0 ' ) 1 / 2 ( 5-17 ) where we have used ( 1-99 ) and taken the constant value of p outside of the integral . The integration over do ' can be ...
Page 360
... becomes 4 / 3z3 which , when put into ( 20-20 ) , gives the induction outside the sphere to be 2 но маз B2o ( z ) = 323 ( 20-22 ) This result becomes more understandable if we express it in terms of the total dipole moment of the sphere ...
... becomes 4 / 3z3 which , when put into ( 20-20 ) , gives the induction outside the sphere to be 2 но маз B2o ( z ) = 323 ( 20-22 ) This result becomes more understandable if we express it in terms of the total dipole moment of the sphere ...
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Ampère's law angle assume axis becomes bound charge boundary conditions bounding surface calculate capacitance capacitor charge density charge distribution charge q circuit conductor consider constant coordinates corresponding Coulomb's law current density curve cylinder defined dielectric dipole direction displacement distance E₁ electric field electromagnetic electrostatic energy equal evaluate example Exercise expression field point flux force free charge free currents frequency function given induction infinitely long integral integrand k₂ Laplace's equation located Lorentz transformation magnetic magnitude material Maxwell's equations normal components obtained origin parallel particle perpendicular plane wave plates point charge polarized position vector potential difference quadrupole quantities radiation radius rectangular region result satisfy scalar scalar potential shown in Figure solenoid sphere spherical tangential components unit vacuum vector potential velocity volume write written xy plane zero Απερ дх Мо