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Unformatted text preview: !0 r 1 0 K 4, K !0 r 2 (c) The free charge density on upper and lower hemispheres are: (= f,ra ) U # (= f , rb ) U # QU Q # and 2 2 2, ra 2, ra (1 0 K ) QU Q Q KQ Q KQ # ; (= f,ra ) L # L 2 # and (= f,rb ) L # L 2 # . 2 2 2 2, rb 2, rb (1 0 K ) 2, ra 2, ra (1 0 K ) 2, rb 2, rb 2 (1 0 K ) % ( K ! 1) & Q % K & % K ! 1 & Q (d) = i,ra # = f,ra (1 ! 1 K ) # ' ( (#' ( 2' 2 ) K * 2, ra ) K 0 1 * ) K 0 1 * 2, ra % ( K ! 1) & Q % K & % K ! 1 & Q ( (#' ( 2' 2 ) K * 2, rb ) K 0 1 * ) K 0 1 * 2, rb (e) There is zero bound charge on the flat surface of the dielectric-air interface, or else that would imply a circumferential electric field, or that the electric field changed as we went around the sphere. EVALUATE:! The charge is not equally distributed over the surface of each conductor. There must be more charge on the lower half, by a factor of K, because the polarization of the dielectric means more free charge is needed on the lower half to produce the same electric field. = i,rb # = f,rb (1 ! 1 K ) # ' 24-26 Chapter 24 24.77. IDENTIFY:! The object is equivalent to two identical capacitors in parallel, where each has the same area A, plate separation d and dielectric with dielectric constant K. !A SET UP:! For each capacitor in the parallel combination, C # 0 . d EXECUTE:! (a) The charge distribution on the plates is shown in Figure 24.77. 2 % ! A & 2(4.2)!0 (0.120 m) (b) C # 2 ' 0 ( # # 2.38 " 10!9 F . 4.5 " 10!4 m )d* EVALUATE:! If two of the plates are separated by both sheets of paper to form a capacitor, C # !0 A 2.38 " 10!9 F , # 2d 4 smaller by a factor of 4 compared to the capacitor in the problem. Figure 24.77 24.78. IDENTIFY:! As in Problem 24.72, the system is equivalent to two capacitors in parallel. One of the capacitors has plate separation d, plate area w( L ! h) and air between the plates. The other has the same plate separation d, plate area wh and dielectric constant K. K !A SET UP:! Define K eff by Ceq # eff 0 , where A #...
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