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Unformatted text preview: ves D D 2 % !0 Ldx & 2 ( K ! 1)!0V L dU # 1 ' dx . ( K ! 1) (V # 2 2D )D * (c) If the charge is kept constant on the plates, then Q # UC %C& !0 LV ( L 0 ( K ! 1) x ) and U # 1 CV 2 # 1 C0V 2 ' ( . 2 2 D ) C0 * & !0 L C0V 2 % ( K ! 1)!0V 2 L ( K ! 1)dx ( and BU # U ! U 0 # ! dx . '1 ! 2 ) DC0 2D * ( K ! 1)!0V 2 L dx , the force is in the opposite direction to the motion dx , meaning that the 2D slab feels a force pushing it out. EVALUATE:! (e) When the plates are connected to the battery, the plates plus slab are not an isolated system. In addition to the work done on the slab by the charges on the plates, energy is also transferred between the battery and ( K ! 1)!0V 2 L . the plates. Comparing the results for dU in part (c) to dU # ! Fdx gives F # 2D IDENTIFY:! C # Q / V . Apply Gauss's law and the relation between potential difference and electric field. r! ! r! ! SET UP:! Each conductor is an equipotential surface. Va ! Vb # H b E U + d r # H b E L + d r , so EU # EL , where these (d) Since dU # ! Fdx # ! 24.76. r0 ra are the fields between the upper and lower hemispheres. The electric field is the same in the air space as in the dielectric. % rr & EXECUTE:! (a) For a normal spherical capacitor with air between the plates, C0 # 4, !0 ' a b ( . The capacitor in ) rb ! ra * this problem is equivalent to two parallel capacitors, CL and CU , each with half the plate area of the normal capacitor. CL # % rr & % rr & % rr & KC0 C # 2, K !0 ' a b ( and CU # 0 # 2, !0 ' a b ( . C # CU 0 CL # 2, !0 (1 0 K ) ' a b ( . 2 2 ) rb ! ra * ) rb ! ra * ) rb ! ra * (b) Using a hemispherical Gaussian surface for each respective half, EL EU QL 4, r 2 QL , so EL # , and # K !0 2, K !0 r 2 2 QU 4, r 2 QU VC K # , so EU # . But QL # VCL and QU # VCU . Also, QL 0 QU # Q . Therefore, QL # 0 # KQU 2 2 !0 2, !0 r 2 and QU # EU # KQ 1 2 Q Q KQ , QL # . This gives EL # # and 2 10 K 10 K 1 0 K 2, K !0 r 1 0 K 4, !0 r 2 Q 1 2 Q # . We do find that EU # EL . 2 1 0 K 2, K...
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