PHYS-102 Summer-08 HOMEWORK-02 Outline of Solutions _____________________________________________________________P19.30The solution is straightforward: the electric flux is given by ()()θΦ===×°=⋅GG42.cos2.00 10 N C 18.0 mcos10.0355 kN mCEEA EA2P19.38(a) GE=0(why is the electric field zero inside the shell?) (b) E=keQr2=8.99×10932.0×10−6()0.2002=7.19 MN CGE=7.19 MN Cradially outwardP19.55F=keq1q2r2: tan=15.060.0=14.0°F1=8.99×10910.0×10−620.1502=40.0 NF3=8.99×10910.0×10−620.600
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This note was uploaded on 09/03/2008 for the course PHYS 102 taught by Professor N/a during the Spring '08 term at Drexel.