Electric Forces and Electric Fields5Answers to Even Numbered Problems 2.135.7110C×4.( )221.91eFkq=aalong the diagonal toward the negative charge 6.92.2510N m−×8.5.08 m 10.( )646.7 N leftF=, ( )1.5157 N rightF=, ( )2111 N leftF−=12.73.8910N at 11.3 below axisx−×°+14., stable if third bead has positive charge 0.634x=d16.1.45 m beyond the –3.00 nC charge 18.(a)72.0010 N C to the right×(b)40.0 N to the left 20.(a)1325.2710m s×(b)55.2710 m s×22.41.6310 N C×directed opposite to the proton’s velocity
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