()()( )22=01.40ˆˆˆˆ3.00 jj3.00 jA j3.00 1.40 1.408.23 N m C.yyyAy=Φ=⋅ −+⋅==⋅(b) The charge is given by ( )1222211enc08.85 10C / N m8.23 N m C7.29 10Cqε−−=Φ=×⋅⋅=×. (c) The electric field can be re-written as0ˆ3.00 jEyE=+GG, where 04.00i6.00jE=−+Gis a constant field which does not contribute to the net flux through the cube. Thus
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This note was uploaded on 09/07/2010 for the course PHYS 244 taught by Professor Ac during the Fall '10 term at The Petroleum Institute.