Question

An electron with speed 2.75×10⁷ m/s is traveling parallel to a uniform electric field of magnitude 1.20×10⁴ N/C .

How far will the electron travel before it stops?

How much time will elapse before it returns to its starting point?

Answer 1

v_i = initial velocity of the electron = 2.75 x 10⁷ m/s

v_f = final velocity when the electron comes to a stop = 0 m/s

q = magnitude of charge on electron = 1.6 x 10^-19 C

m = mass of electron = 9.1 x 10^-31 kg

E = magnitude of electric field = 1.20 x 10⁴ N/C = 12000 N/C

a = magnitude of acceleration of electron

d = distance travelled by electron before it comes to stop = ?

magnitude of acceleration of electron is given as

a = qE/m

a = (1.6 x 10^-19) (12000)/(9.1 x 10^-31)

a = 2.11 x 10¹⁵ m/s²

When the electron is slowing down, acceleration is opposite to the velocity and hence taken as negative

v²_f = v²_i + 2ad

0² = (2.75 x 10⁷)² + 2 (- 2.11 x 10¹⁵) d

d = 0.18 m

when the electron comes to its starting point, it velocity has same magnitude but in opposite direction.

v_i = initial velocity of the electron = 2.75 x 10⁷ m/s

v_f = final velocity when the electron comes to its starting position = - 2.75 x 10⁷ m/s

a = - 2.11 x 10¹⁵ m/s²

t = time taken

Using the equation

v_f = v_i + at

- 2.75 x 10⁷ = 2.75 x 10⁷ + (- 2.11 x 10¹⁵) t

t = 2.61 x 10^-8 sec