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 $\mathrm{\mu F}$ is connected as shown in figure. Then
 

Q = CV = C (Bvl) = 10 x ${10}^{-6}$ x 4 x 2 x 1 = 80 $\mathrm{\mu C}$. According 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.