In the given circuit diagram (figure), switch SW  is shifted from position 1 to position 2. Then

Initially, when the switch is closed on position 1, the capacitor C is connected in series with batteries E1  and  E2. From KVL, we haveQiC−E2+E1=0or Qi=(E2−E1)C..........(i)Depending upon the sign of (E2−E1),charge Qion the left plate may be positive if  (E2>E1) ,  or  negative (if  E2