Electrostatics : Electron in electric field

Question :

A electron of mass m and charge -q is at rest at origin on x-axis. It is subjected to an electric field which is given below

E=E₀sinωti

(1) Find out the position of the electron as a function of time

a)$x=-\frac{q{E}_0}{m{\omega }^2}\sin \mathrm{\omega t}$

b)$x=\frac{q{E}_0}{\mathrm{m\omega }}t-\frac{q{E}_0}{m{\omega }^2}\sin \mathrm{\omega t}$

c)$x=-\frac{q{E}_0}{\mathrm{m\omega }}t+\frac{q{E}_0}{m{\omega }^2}\sin \mathrm{\omega t}$

d)$x=-\frac{q{E}_0}{\mathrm{m\omega }}t-\frac{q{E}_0}{m{\omega }^2}\sin \mathrm{\omega t}$

(2) Find out the velocity of the electron as a function of time

a)v$=\frac{q{E}_0}{\mathrm{m\omega }}(\cos \mathrm{\omega t}-1)$

b)v$=\frac{q{E}_0}{\mathrm{m\omega }}(\cos \mathrm{\omega t}+1)$

c)v$=\frac{q{E}_0}{\mathrm{m\omega }}(\sin \mathrm{\omega t}-1)$

d)v$=\frac{q{E}_0}{\mathrm{m\omega }}(\sin \mathrm{\omega t}+1)$

Solution

The force on electron due to electric field

F=-qE

F=-qE₀sinωti

Acceleration

$a=\frac{-q{E}_0}{m}\sin \mathrm{\omega t}$i

or

$\frac{\mathrm{dv}}{\mathrm{dt}}$$=\frac{-q{E}_0}{m}\sin \mathrm{\omega t}$

Integrating we get

$v=\frac{q{E}_0}{\mathrm{m\omega }}(\cos \mathrm{\omega t}-1)+a$

Where a is integration constants

Since v=0 at t=0

$v=\frac{q{E}_0}{\mathrm{m\omega }}(\cos \mathrm{\omega t}-1)$

Now again

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{q{E}_0}{\mathrm{m\omega }}(\cos \mathrm{\omega t}-1)$

Integrating we get

$x=-\frac{q{E}_0}{\mathrm{m\omega }}t-\frac{q{E}_0}{m{\omega }^2}\sin \mathrm{\omega t}+b$

Where b is integration constants

Now x=0 at t=0

So

x$=-\frac{q{E}_0}{\mathrm{m\omega }}t-\frac{q{E}_0}{m{\omega }^2}\sin \mathrm{\omega t}$