A particle of charge q, mass m is moving in an electric field `vecE= EhatiNC^(-1)` and magnetic field `vecB =BhatkT` . It follows a trajectory from P to Q as shown Velocity at P is `Vhati` & that at Q is `- 2Vhatj`, find
a) electric field strength
b) power developed at P by electric field
c) rate of work done at by both fields at Q

Explanation

(a) Work done by magnetic field is zero `becausevecFbotvecV` only `vecE` does work on charged particle
From work energy theorem
`DeltaKE=WD` by `vecE`
`1/2m(v^(2)-u^(2))=vecEq.vecS`
`1/2m(4v^(2)-v^(2))=q(Ehati.2ahati)`
`1/2m3v^(2)=2Eq.a`

b) power developed by `vecE` at P is
`vecE.vecV=q(Ehati.vhati)=q(vE)=qv((3mv^(2))/(4aq))rArrP=(3mv^(3))/(4a)`
c) Power developed by both fields is equal to power developed by `vecE` alone (`because` Work done by `vecB` is 0)
`vecP_(0)=vecF.vecV_(0)=q[(vecEhati.(-2Vhatj)]=0`