Asked by iowa
                An electron and a proton have charges of opposite sign, but the magnitude of their charges has the same value of 1.60  10-19 C. Suppose the electron and proton in a hydrogen atom are separated by a distance of 4.30  10-11 m. What are the magnitude and direction of the electrostatic force exerted on the electron by the proton?
            
            
        Answers
                    Answered by
            Steve
            
    clearly the force is attractive, pointing toward the proton.
F = kqq/r^2
= (8.99*10^9)(1.60*10^-19)^2/(4.30*10^-11)^2
= 1.24*10^-7 N
 
    
F = kqq/r^2
= (8.99*10^9)(1.60*10^-19)^2/(4.30*10^-11)^2
= 1.24*10^-7 N
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