1. (a) We note that the electric field points leftward at both points. UsingGGFqE=0, and orienting our xaxis rightward (so ˆi points right in the figure), we find ()1918Nˆˆ1.6 10C40i( 6.4 10N)iCF−−⎛⎞=+ ×−=− ×⎜⎟⎝⎠Gwhich means the magnitude of the force on the proton is 6.4 × 10–18N and its direction
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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.