An electron moving with a speed of 5 times 10 | vedclass

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(C) The electric force acting on the electron is $F = qE = ma$. Thus,the retardation $a = \frac{qE}{m}$. Substituting the given values: $a = \frac{1.6

Vedclass · Accepted Answer

$7 \, cm$

Vedclass · Answer

(C) The electric force acting on the electron is $F = qE = ma$. Thus,the retardation $a = \frac{qE}{m}$. Substituting the given values: $a = \frac{1.6 	imes 10^{-19} 	imes 1 	imes 10^3}{9 	imes 10^{-31}} = \frac{1.6}{9} 	imes 10^{15} \, m/s^2$. Given initial velocity $u = 5 	imes 10^6 \, m/s$ and final velocity $v = 0$. Using the kinematic equation $v^2 = u^2 - 2as$,we find the distance $s = \frac{u^2}{2a}$. $s = \frac{(5 	imes 10^6)^2}{2 	imes (\frac{1.6}{9} 	imes 10^{15})} = \frac{25 	imes 10^{12} 	imes 9}{3.2 	imes 10^{15}} = \frac{225}{3.2} 	imes 10^{-3} \approx 70.3 	imes 10^{-3} \, m = 7.03 	imes 10^{-2} \, m \approx 7 \, cm$.

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