Question

A particle of mass m and charge q is placed in an electric field e which varies with time as equal to

Answer 1

Answer:

Equation of motion of the particle is

mdvdt=qE0sin(ωt) mdvdt=qE0sin(ωt)  

which upon integrating gives,  

m∫dv=qE0∫sin(ωt)dtv=−qE0mωcos(ωt) +C1 m∫dv=qE0∫sin(ωt)dtv=-qE0mωcos(ωt) +C1  

where C1 is the constant of integration.

so ,

dxdt=−qE0mωcos(ωt) +C1 dxdt=-qE0mωcos(ωt) +C1  

which , upon integration again , gives  

∫dx=−qE0mω∫(cos(ωt) +C1)dtx=−qE0mω2sin(ωt) +C1t+C2∫dx=−qE0mω∫(cos(ωt) +C1)dtx=-qE0mω2sin(ωt) +C1t+C2

The first term on the right hand provides the simple harmonic part ,

So, the amplitude of the simple harmonic part of the solution is  

qE0mω2

Explanation: