Question

An electron moving to the right at 1.0% the speed of

lightenters a uniform electric field parallel to its direction
ofmotion. If the electron is to be brought to rest in the spaceof
4.0 cm, (a) what direction is required for the electric field,and
(b) what is the strength of the field?

Answer 1

(a) In the direction of velocity

(b) E = - 639.84 N/C

Solution:

As per the question:

Speed of the electron,
`v_(e)` = 0.1% c

where

c =
`3* 10^(8)\ m/s`

Thus

`v_(e) = 3* 10^(6)\ m/s`

Distance, x = 4.0 cm

Now,

(a) The direction of the electric field is the same as that of the velocity.

(b) The electric field strength can be calculated as:

By Kinematics eqn:

`v'^(2) = v_(e)^(2) + 2ax`

`0 = (3* 10^(6))^(2) + 2a* 0.04`

`a = -1.125* 10^(14)\ m/s^(2)`

Electric field strength can be calculated as:

`E = (F)/(q)`

Also,

F = ma

m = mass of electron =
`9.1* 10^(- 31)\ kg`

q = charge on electron =
`1.6* 10^(- 19)\ C`

Thus

`E = (ma)/(q) = (9.1* 10^(- 31)* - 1.125* 10^(14))/(1.6* 10^(- 19))`

E = - 639.84 N/C