An electron is accelerated by a constant electric field of magnitude 295 N/C.(a) Find the acceleration of the electron.m/s2(b) Use the equations of motion with constant acceleration to find the electron's speed after 1.25 x 108 s, assuming it starts fromrest.m/s

Answered: An electron is accelerated by a…

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An electron is accelerated by a constant electric field of magnitude 295 N/C.
(a) Find the acceleration of the electron.
m/s2
(b) Use the equations of motion with constant acceleration to find the electron's speed after 1.25 x 108 s, assuming it starts from
rest.
m/s

Expert Solution

Step 1

(a)Electric field of an electron,(E)=(-)295 N/C -----(1) [Given] [Negative sign just indicates the direction of field]

mass of an electron,(m)=9.11 x 10^-31 kg -----(2) [Given]

Charge on electron,(q)=(-)1.6 x 10^-19 C -----(3) [Given] [Negative sign just indicates the direction of charge]

Let us consider the acceleration of the electron be (a) in meter/second² respectively.

We know, $E = \frac{m a}{q}$ ------(4)

Putting the value of E,m & q from (1),(2) & (3) in (4) we get,

$- 295 N / C = \frac{9.11 \times 10^{- 31} k g \times a}{- 1.6 \times 10^{- 19} C}$

$295 N / C \times 1.6 \times 10^{- 19} C = 9.11 \times 10^{- 31} k g \times a$

$9.11 \times 10^{- 31} k g \times a = 295 N / C \times 1.6 \times 10^{- 19} C$

$9.11 \times 10^{- 31} k g \times a = 295 \times 1.6 \times 10^{- 19} N$

$a = \frac{295 \times 1.6 \times 10^{- 19} N}{9.11 \times 10^{- 31} k g}$

$a = \frac{295 \times 1.6 \times 10^{- 19} k g - m / s e c^{2}}{9.11 \times 10^{- 31} k g}$ [Using 1 N=1 kg-m/sec²]

$a = (\frac{295 \times 1.6 \times 10^{- 19}}{9.11 \times 10^{- 31}}) m / s e c^{2}$

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An electron is accelerated by a constant electric field of magnitude 295 N/C. (a) Find the acceleration of the electron. m/s² (b) Use the equations of motion with constant acceleration to find the electron's speed after 1.25 x 108 s, assuming it starts from rest. m/s