A conducting rod PQ of length L = 1.0 m is moving with a uniform speed v = 2 m/s in a uniform magnetic field  B = 4.0 T directed into the paper. A capacitor of capacity C = 10 μF is connected as shown in figure. Then

Q=CV=C(Bvl)=10×10−6×4×2×1=80μCAccording to Fleming's right hand rule induced current flows from Q to P. Hence P is at higher potential and Q is at lower potential. Therefore A is positively charged and B is negatively charged.