An electron falls through a distance of 1.5 , | vedclass

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(B) The electric force on the electron is given by $F = qE$,where $q = 1.6 	imes 10^{-19} \, C$ and $E = 2.0 	imes 10^4 \, N/C$.$F = 1.6 	imes 10^{

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$2.9 	imes 10^{-9} \, s$

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(B) The electric force on the electron is given by $F = qE$,where $q = 1.6 	imes 10^{-19} \, C$ and $E = 2.0 	imes 10^4 \, N/C$.$F = 1.6 	imes 10^{-19} 	imes 2.0 	imes 10^4 = 3.2 	imes 10^{-15} \, N$.The acceleration $a$ of the electron is $a = \frac{F}{m_e} = \frac{3.2 	imes 10^{-15}}{9.1 	imes 10^{-31}} \, m/s^2$.Using the kinematic equation for distance $S = \frac{1}{2}at^2$,where $S = 1.5 \, cm = 1.5 	imes 10^{-2} \, m$:$t = \sqrt{\frac{2S}{a}} = \sqrt{\frac{2 	imes 1.5 	imes 10^{-2} 	imes 9.1 	imes 10^{-31}}{3.2 	imes 10^{-15}}}$.$t = \sqrt{\frac{3.0 	imes 9.1 	imes 10^{-43}}{3.2 	imes 10^{-15}}} = \sqrt{8.53 	imes 10^{-28}} \approx 2.9 	imes 10^{-9} \, s$.

An electron falls through a distance of $1.5 \, cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \, N/C$ as shown in the figure. The time taken by the electron to fall through this distance is ($m_e = 9.1 \times 10^{-31} \, kg$,neglect gravity).

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