15. (a) The electron ecis a distance r= z= 0.020 m away. Thus, 922196220(8.99 10 N m C )(1.60 10C)3.60 10N/C4(0.020 m)CeErπε−−×⋅×===×. (b) The horizontal components of the individual fields (due to the two escharges) cancel, and the vertical components add to give 219s,net3/2223/20622(8.99 10 N m C )(1.6 10C)(0.020 m)4()[(0.020 m)(0.020 m) ]2.55 10N/C .ezERz−−×++=×(c) Calculation similar to that shown in part (a) now leads to a stronger field 43.60 10 N/CcE−from the central charge. (d) The field due to the side charges may be obtained from calculation similar to that
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This note was uploaded on 06/07/2010 for the course PHYS 344 taught by Professor Bb during the Spring '10 term at The Petroleum Institute.