An insulating rod with linear charge density of 40 μC/m and linear mass density of 0.1 kg/m is released from rest in a uniform electric field E = 100 V/m directed perpendicular to the rod. If rod is released and after travelling 2m down the field, speed of rod is found to be (v/10) m/s, then value of v is ______

Taking potential Vinitial = 0 at initial position of rod, then potential after travelling distance 'd’ is V=-EdCharge on rod = λLPotential energy of rod-field system initially QVinitial=0Potential energy of rod-field system finally=−λLEd (using, U=qV)Now using energy conservation, (KE+PE)initial =(KE+PE)final  We get, 0+0=12mv02−λLEd⇒ v02=2λLEdm=2λLEdμL=2λEdμ⇒ v0=2λEdμ=2×40×10−6×100×20.1 =0.400m/s=410m/s So,  v=4