Question #317988

Two neutrally-charged bodies are separated by 2cm. Electrons are moved from one body and placed on the second body until a force of 1.7x10-7 N is generated between them. How many electrons were transferred between the bodies? 


1
Expert's answer
2022-03-25T09:10:59-0400

The Coulomb's law says

F=kq1q2r2F=k\frac{|q_1||q_2|}{r^2}

The charges of bodies have the same magnitude, so

q=Fr2k=1.71070.029109=6.11010Cq=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{1.7*10^{-7}*0.02}{9*10^9}}=6.1*10^{-10}\:\rm C

The number of electrons transferred between the bodies

N=qe=6.110101.61019=3.8109N=\frac{q}{e}=\frac{6.1*10^{-10}}{1.6*10^{-19}}=3.8*10^{9}


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