hw07a - Physics 505 Homework Assignment #7 Solutions...

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Physics 505 Fall 2005 Homework Assignment #7 — Solutions Textbook problems: Ch. 4: 4.10 Ch. 5: 5.3, 5.6, 5.7 4.10 Two concentric conducting spheres of inner and outer radii a and b , respectively, carry charges ± Q . The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant ±/± 0 0, as shown in the figure. a Q - Q + b a ) Find the electric field everywhere between the spheres. This is a somewhat curious problem. It should be obvious that without any dielectric the electric field between the spheres would be radial ~ E = Q 4 π± 0 ˆ r r 2 We cannot expect this to be unmodified by the dielectric. However, we note that the radial electric field is tangential to the interface between the dielectric and empty region. Thus the tangential matching condition E k 1 = E k 2 is automatically satisfied. At the same time there is no perpendicular component to the interface, so there is nothing to worry about for the D 1 = D 2 matching condition. This suggests that we guess a solution of the radial form ~ E = A ˆ r r 2 where A is a constant to be determined. This guess is perhaps not completely obvious because one may have expected the field lines to bend into or out of the dielectric region. However, we could also recall that parallel fields do not get bent across the dielectric interface. We may use the integral form of Gauss’ law in a medium to determine the above constant A I ~ D · ˆ n da = Q ± 0 A r 2 (2 πr 2 ) + ±A r 2 (2 πr 2 ) = Q
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or A = Q/ 2 π ( ± + ± 0 ). Hence ~ E = Q 2 π ( ± + ± 0 ) ˆ r r 2 Note that 1 2 ( ± + ± 0 ) may be viewed as the average permittivity in the volume between the spheres. b ) Calculate the surface-charge distribution on the inner sphere. The surface-charge density is given by σ = D ± ± r = a where either D = ± 0 E or D = ±E depending on region. This gives σ = ± ± + ± 0 Q 2 πa 2 ; dielectric side ± 0 ± + ± 0 Q 2 πa 2 ; empty side (1) Note that the total charge obtained by integrating σ over the surface of the inner sphere gives Q as expected. c ) Calculate the polarization-charge density induced on the surface of the dielectric at r = a . The polarization charge density is given by ρ pol = -∇· ~ P where ~ P = ± 0 χ e ~ E = ( ± - ± 0 ) ~ E . Since the surface of the dielectric at r = a is against the inner sphere, we can take the polarization to be zero inside the metal (‘outside’ the dielectric). Gauss’ law in this case gives σ pol = - P ± ± r = a = - ( ± - ± 0 ) E ± ± r = a = - ± - ± 0 ± + ± 0 Q 2 πa 2 Note that when this is combined with (1), the total (free and polarization) charge
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This note was uploaded on 11/21/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.

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hw07a - Physics 505 Homework Assignment #7 Solutions...

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