Considering the activity from `P to Q` ( Horizontal)
` u_(1) = v, v_(1) = 2v, s_(1) = 2a , Acc = A`
rArr ` 4v^(2)- v^(2) = 2A(2a)`
rArr `A = (3v^(2))/(4a)`
Force acting in the horizontal direction is
` F = q ^(E) = mA`
rArr `E = (mA)/(q) = 3/4[ (mv^(2))/(qa)]`
Rate of doing work A `P`
power ` = Fxxv = mAxxv = 3/4[(mv^(3)/(a)]`
Rate of doing work by the magnetic field is throughout zero. The rate of doing work by eletric field is zero at `Q` . Because at `Q` , the angle between force due to electric field and displacement is zero .
