The electric field intensity just sufficient | vedclass

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Question

(A) To balance the gravitational force on an electron,the upward electric force must be equal to the downward gravitational force.Let $E$ be the elect

Vedclass · Accepted Answer

$-5.6 	imes 10^{-11} \ N/C$

Vedclass · Answer

(A) To balance the gravitational force on an electron,the upward electric force must be equal to the downward gravitational force.Let $E$ be the electric field intensity. The electric force on the electron is $F_e = qE = -eE$ (since the charge of an electron is negative).The gravitational force on the electron is $F_g = mg$.For equilibrium,the forces must balance: $F_e + F_g = 0$,which implies $eE = mg$ in magnitude,or $-eE = mg$ considering the direction.Thus,$E = -\frac{mg}{e}$.Substituting the given values:$E = -\frac{9.1 	imes 10^{-31} \ kg 	imes 10 \ m/s^2}{1.6 	imes 10^{-19} \ C}$$E = -\frac{9.1 	imes 10^{-30}}{1.6 	imes 10^{-19}} \ N/C$$E = -5.6875 	imes 10^{-11} \ N/C \approx -5.6 	imes 10^{-11} \ N/C$.

The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (Given: mass of an electron $m = 9.1 \times 10^{-31} \ kg$,charge of an electron $e = 1.6 \times 10^{-19} \ C$,and acceleration due to gravity $g = 10 \ m/s^2$.)

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