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(C) The force experienced by an electron in an electric field is given by $F = qE$,where $q = e$ is the charge of the electron and $E$ is the electric

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$5.5 	imes 10^{-11} \ N/C$

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(C) The force experienced by an electron in an electric field is given by $F = qE$,where $q = e$ is the charge of the electron and $E$ is the electric field intensity.Given that the force is equal to the weight of the electron,we have $F = mg$,where $m$ is the mass of the electron $(m \approx 9.1 	imes 10^{-31} \ kg)$ and $g$ is the acceleration due to gravity $(g \approx 9.8 \ m/s^2)$.Equating the two,we get $eE = mg$,which implies $E = \frac{mg}{e}$.Substituting the values: $E = \frac{9.1 	imes 10^{-31} \ kg 	imes 9.8 \ m/s^2}{1.6 	imes 10^{-19} \ C}$.$E \approx \frac{89.18 	imes 10^{-31}}{1.6 	imes 10^{-19}} \ N/C \approx 5.57 	imes 10^{-11} \ N/C$.Rounding to the nearest given option,the intensity of the field is $5.5 	imes 10^{-11} \ N/C$.

An electron experiences a force equal to its weight when placed in an electric field. The intensity of the field will be

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