In the circuit shown, switch S2 is closed first and is kept closed for a long time. Now S1 is closed. Just after that instant, the current through S1 is

Just before S1 is closed, the potential difference across capacitor 2 is 2ε.Just after S1 is closed, the potential differences across capacitors 1 and 2 arc 0 and 2ε respectively.If we take potential of C and D to be zero, then VA = ε,   VB = 2εSince VB > VA, so current will flow i=2ε−εR1 =εR1 towards  left