Question

An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. If the distance the electron travels through the field is 1.0 cm, how far is it deviated (Y) in its path when it emerges from the electric field? (me = 9.1 ´ 10 - 31 kg, e = 1.6 ´ 10 - 19 C)

Answer 1

Answer:

The deviation in path is $4.39 \times 10^{-3}$

Explanation:

Given:

Velocity $v = 1 \times 10^{6}$ $\frac{m}{s}$

Electric field $E = 500$ $\frac{V}{m}$

Distance $x = 1 \times 10^{-2}$ m

Mass of electron $m = 9.1 \times 10^{-31}$ kg

Charge of electron $q = 1.6 \times 10^{-19}$ C

Time taken to travel distance,

    $t = \frac{x}{v}$

    $t = \frac{1 \times 10^{-2} }{1 \times 10^{6} }$

    $t = 10^{-8}$ sec

Acceleration is given by,

  $F = qE$

$ma = qE$

   $a = \frac{qE}{m}$

   $a = \frac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31} }$

   $a = 8.77 \times 10^{13}$ $\frac{m}{s^{2} }$

For finding the distance, we use kinematics equations.

   $y = vt + \frac{1}{2} at^{2}$

Where $v = 0$ because here initial velocity zero

   $y = \frac{1}{2} at^{2}$

   $y = \frac{1}{2} \times 8.77 \times 10^{13 } \times (10^{-2} )^{2}$

   $y = 4.39 \times 10^{-3}$ m

Therefore, the deviation in path is $4.39 \times 10^{-3}$