Respuesta :
The force of a charge in an electric field is:
[tex]\vec{F}_e=q\vec{E}[/tex]In this case we know the electric field is:
[tex]\vec{E}=104\hat{j}[/tex]and that the charge is that of the electron, then we have:
[tex]\begin{gathered} \vec{F}_e=-1.6\times10^{-19}(104\hat{j}) \\ \vec{F}_e=-1.664\times10^{-17}\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the force is
[tex]1.664\times10^{-17}\text{ N}[/tex]and in points down.
The weight of the electron is:
[tex]\begin{gathered} W=1.67\times10^{-27}(9.98) \\ W=1.6366\times10^{-26} \end{gathered}[/tex]Making the quotient between the force we have:
[tex]\frac{1.664\times10^{-17}}{1.6366\times10^{-26}}=1.02\times10^9[/tex]Therefore, the electric force is approximately 1e9 times the weight.
