Answer:
(a) In the direction of velocity
(b) E = - 639.84 N/C
Solution:
As per the question:
Speed of the electron, [tex]v_{e}[/tex] = 0.1% c
where
c = [tex]3\times 10^{8}\ m/s[/tex]
Thus
[tex]v_{e} = 3\times 10^{6}\ m/s[/tex]
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:
[tex]v'^{2} = v_{e}^{2} + 2ax[/tex]
[tex]0 = (3\times 10^{6})^{2} + 2a\times 0.04 [/tex]
[tex]a = -1.125\times 10^{14}\ m/s^{2}[/tex]
Electric field strength can be calculated as:
[tex]E = \frac{F}{q}[/tex]
Also,
F = ma
m = mass of electron = [tex]9.1\times 10^{- 31}\ kg[/tex]
q = charge on electron = [tex]1.6\times 10^{- 19}\ C[/tex]
Thus
[tex]E = \frac{ma}{q} = \frac{9.1\times 10^{- 31}\times - 1.125\times 10^{14}}{1.6\times 10^{- 19}}[/tex]
E = - 639.84 N/C
