Question

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

Answer 1

Answer:

(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 = $3\times 10^{8}\ m/s$

Thus

$v_{e} = 3\times 10^{6}\ m/s$

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:

$v'^{2} = v_{e}^{2} + 2ax$

$0 = (3\times 10^{6})^{2} + 2a\times 0.04$

$a = -1.125\times 10^{14}\ m/s^{2}$

Electric field strength can be calculated as:

$E = \frac{F}{q}$

Also,

F = ma

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

q = charge on electron = $1.6\times 10^{- 19}\ C$

Thus

$E = \frac{ma}{q} = \frac{9.1\times 10^{- 31}\times - 1.125\times 10^{14}}{1.6\times 10^{- 19}}$

E = - 639.84 N/C