Question

A particle of mass 0.001 kg and charge 1 C is initially at rest and at time, t= 0 the particle comes under the influence of electric field $E={{E}_{0}}\sin (wt)\widehat{i}$where ${{E}_{0}}=1N/C\ And w={{10}^{3}}rad/s$. Consider the effect of only the electrical force on the particle. Then the maximum speed attained by it at subsequent times is?

Answer 1

Here given a charged particle the magnitude of charge is 1 C, so, this is a positive charge. It enters into a region where the electric field is present. Electric field will exert electric force and since charge is in motion its trajectory may change.

Complete step by step answer:
Mass, m= 0.001 kg
Charge, q= 1 C
Electric field is $E={{E}_{0}}\sin (wt)\widehat{i}$ where ${{E}_{0}}=1N/C\ And w={{10}^{3}}rad/s$
We know magnitude of electric force is given by, F= qE
$F=q{{E}_{0}}\sin (wt)$
We need to find the total time for which force is experienced, it acts from time, t= 0 to t=$\dfrac{\pi }{w}$
$ma=q{{E}_{0}}\sin (wt) \\\ \implies a=\dfrac{q{{E}_{0}}\sin (wt)}{m} \\\ \implies \dfrac{dv}{dt}=\dfrac{q{{E}_{0}}\sin (wt)}{m} \\\ \implies \int\limits_{0}^{v}{dv}=\int\limits_{0}^{\dfrac{\pi }{w}}{\dfrac{q{{E}_{0}}\sin (wt)}{m}dt} \\\ \implies v=\dfrac{q{{E}_{0}}}{m}\int\limits_{0}^{\dfrac{\pi }{w}}{\sin wtdt} \\\ \implies v=\dfrac{q{{E}_{0}}}{wm}[-\cos wt]_{0}^{\dfrac{\pi }{w}} \\\ \implies v=\dfrac{q{{E}_{0}}}{wm}[-\cos \pi -\cos 0] \\\ \therefore v=\dfrac{2q{{E}_{0}}}{wm} \\\$
putting the values,
$v=\dfrac{1\times 1}{0.001\times 1000}\times 2=2m/s$

So, the value of the velocity comes out to be 2 m/s.

Additional Information:
A velocity selector is a region in which there are uniform electric fields and a uniform magnetic field. The fields are perpendicular to one another, and perpendicular to the velocity of the charged particle

Note:
We were not given the value of time, so we had taken it to be a time period because the electric field is sinusoidal and it one complete cycle takes total time = time period. Had there been magnetic fields too, we would then have to take into account the magnetic force also.