In a uniform electric field of E = 1 times 10 | vedclass

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Question

(A) According to the work-energy theorem,the work done by the electric field is equal to the change in the kinetic energy of the electron.Work done $W

Vedclass · Accepted Answer

$8.5 	imes 10^{6} 	ext{ m s}^{-1}$

Vedclass · Answer

(A) According to the work-energy theorem,the work done by the electric field is equal to the change in the kinetic energy of the electron.Work done $W = qEx = \frac{1}{2}mv^2$.Rearranging for velocity $v$,we get $v = \sqrt{\frac{2qEx}{m}} = \sqrt{2 \left(\frac{e}{m}ight) Ex}$.Given values: $\frac{e}{m} = 1.8 	imes 10^{11} 	ext{ C kg}^{-1}$,$E = 1 	imes 10^{4} 	ext{ N C}^{-1}$,and $x = 2 	imes 10^{-2} 	ext{ m}$.Substituting these values:$v = \sqrt{2 	imes (1.8 	imes 10^{11}) 	imes (1 	imes 10^{4}) 	imes (2 	imes 10^{-2})}$$v = \sqrt{7.2 	imes 10^{13}} = \sqrt{72 	imes 10^{12}}$$v \approx 8.485 	imes 10^{6} 	ext{ m s}^{-1} \approx 8.5 	imes 10^{6} 	ext{ m s}^{-1}$.

In a uniform electric field of $E = 1 \times 10^{4} \text{ N C}^{-1}$,an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $2 \times 10^{-2} \text{ m}$ is nearly (given $\frac{e}{m} \text{ of electron } \approx 1.8 \times 10^{11} \text{ C kg}^{-1}$):

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