Question

9.26 * 10^ 11 is wrong . Submitted this question 2 times both answers said 9.26 * 10^11. It is not correct

An electron with speed vo-1.29x106 m/s is traveling parallel to and along the same direction as an electric field of magnitude E 7.93x103 N/C. (a) How far will the electron travel before it stops? 9810 m You are correet. Computer's answer now shown above Your receipt is 160-2359 0 Pieskus Ines b How much time will elapse before it returns to its starting point? 9.26 e-10 s

Answer 1

(b) The force on the electron is given by,

$F = qE \Rightarrow F = (-1.6 \times 10^{-19})(7.93 \times 10^3) = -1.27 \times 10^{-15} \, N$

Therefore the acceleration of the electron is,

$F = ma \Rightarrow ma = -1.27 \times 10^{-15} \, N$

$\Rightarrow a = \frac{-1.27 \times 10^{-15} \, N}{9.1 \times 10^{-31} \, kg} = -1.4 \times 10^{15} \, m/s^2$

The distance traveled by the electron before it stops is 5.98 x 10^(-4) m. The time taken to travel the distance is given by, ie. the time taken for the electron to get at a complete stop is,

$v = u + at \Rightarrow 0 = 1.29 \times 10^6 - 1.4 \times 10^{15}t$

$\Rightarrow t = \frac{1.29 \times 10^6}{1.4 \times 10^{15}} = 9.21 \times 10^{-10} \,s$

Since the motion is symmetric, we can say that this same time will be taken for the electron to come back to its initial starting position. Therefore the time taken to come back to starting point will be twice of this at it will be,

$T = 2(9.21 \times 10^{-10}) = 1.84 \times 10^{-9} \,s$