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

Hint: 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.