Energy stored in capacitor when connected or disconnected from battery

[FollowTheFez]

Homework Statement

A capacitor is charged by connecting it to a suitable battery.

(a) Does the amount of energy stored in the capacitor increase, decrease, or remain the same if a dielectric is inserted while the capacitor is connected to the battery? Explain your answer.

(b) Does the amount of energy stored in the capacitor increase, decrease, or remain the same if a dielectric is is inserted after the battery is disconnected? Explain your answer.

Homework Equations

C=kε₀ $\frac{A}{d}$

E=$\frac{1}{2}$QV

Q=CV

The Attempt at a Solution

I know that when a dielectric is inserted into a capacitor the capacitance increases by a factor of k since C=kε₀ $\frac{A}{d}$

I am getting confused as to what happens when the capacitor is/is not connected to the battery. My textbook doesn't explain it very well.
Can someone please tell me if I am correct?

For a) While connected to battery, the capacitor is charging or is already fully charged.
So Q=max and C=max. But C also depends on the dielectric by:
C=kε₀ $\frac{A}{d}$
Therefore, regardless of what the max Q and C values are, if a dielectric is inserted into the capacitor, the capacitance will increase causing the Q to increase.
And since E=$\frac{1}{2}$QV , if Q increases then the amount of energy stored in the capacitor will increase.

For b) The capacitor is disconnected from battery and is therefore discharging. The principals explained in (a) still apply and therefore if a dielectric is inserted E will increase. However the capacitor is discharging to E is continually decreasing but it is larger than what it would be if no dielectric were present.

 

[voko]

When a capacitor is connected to a battery and is fully charged, what is V? Does it change if the capacitance changes? Why?

 

[gneill]

For b) The capacitor is disconnected from battery and is therefore discharging. The principals explained in (a) still apply and therefore if a dielectric is inserted E will increase. However the capacitor is discharging to E is continually decreasing but it is larger than what it would be if no dielectric were present.

If the capacitor is disconnected (isolated), where is it discharging to?