An electron with mass 9.1 times 10^{-31} kg | vedclass

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(B) Given: Mass $m = 9.1 	imes 10^{-31} \ kg$,Charge $q = 1.6 	imes 10^{-19} \ C$,Electric field $E = 1 	imes 10^6 \ V/m$,Final velocity $v = \frac

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$1.7 	imes 10^{-10} \ s$

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(B) Given: Mass $m = 9.1 	imes 10^{-31} \ kg$,Charge $q = 1.6 	imes 10^{-19} \ C$,Electric field $E = 1 	imes 10^6 \ V/m$,Final velocity $v = \frac{c}{10} = \frac{3 	imes 10^8}{10} = 3 	imes 10^7 \ m/s$.Using Newton's second law and the definition of electric force: $F = qE = ma$.Acceleration $a = \frac{qE}{m} = \frac{(1.6 	imes 10^{-19}) 	imes (1 	imes 10^6)}{9.1 	imes 10^{-31}} = \frac{1.6}{9.1} 	imes 10^{18} \ m/s^2$.Using the equation of motion $v = u + at$,where initial velocity $u = 0$:$v = at \implies t = \frac{v}{a} = \frac{3 	imes 10^7}{\frac{1.6}{9.1} 	imes 10^{18}} = \frac{3 	imes 9.1}{1.6} 	imes 10^{-11} \ s$.$t = 17.0625 	imes 10^{-11} \ s \approx 1.7 	imes 10^{-10} \ s$.

An electron with mass $9.1 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is placed in an electric field of $1 \times 10^6 \ V/m$. How much time will it take for the electron to reach a velocity equal to $1/10$th of the speed of light?

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