So Q's potential must be negative as the current flow from A to Q.
After S is closed the p.d. of A and Q increases with the e.m.f. supply of the battery.
So potential of Q must be more negative(decreases) in order for the increase of p.d.
After S is closed. the p.d. between P and A decreases. As potential of A is zero, the potential of P should be less positive (decrease) in order for the decreased p.d.
Are my explanations correct? If not, please explain. Thanks for your help.
After S
更新1:
The correct ans is E