Question

capacitor question that is too long to paste:
http://i.imgur.com/Cwvh045.png

Answer 1

i assume you know \[Q=CV\]\[E=\frac{ 1 }{ 2 }CV ^{2}=\frac{ 1 }{ 2 }QV\] right! where E is the energy stored in the electric field . also\[C~ is~ proportional~t _{O}~~Area~of~plates/separation~of ~plates\]

In the first situation after the capacitor is charged and the battery is removed you must realize that Q cannot change, ie it it fixed no matter separation.

In the second situation the battery being left connected maintain the voltage no matter the separation.