A capacitor is charged and then disconnected from the battery. When you insert a dielectric into a capacitor while the capactior is charged but the disconnected from the battery, does the energry stored in the capactior increase or decrease? Please explain and use formulas

There is a parallel plate capacitor connected to a battery and a dielectric material is partially inserted between the capacitor plates. Which one of the below statements is wrong?

Consider the capacitor below, that is not connected to anything. The capacitor has a dielectric between its plates initially. The dielectric is slowly pulled out from between the plates. As that happens, how will the energy stored in the capacitor change? V1+ C1 O It will increase. O It will decrease. O It will stay the same.

plz answer, thankss

A parallel-plate capacitor of capacitance C0 = ε0A0/d0 , is connected to a battery. When the capacitor is fully charged, it is disconnected from the battery (initial state). We push the plates closer together, until d = d0/2 (final state).  a) How does the energy stored in the capacitor change (increase or decrease)? Justify your answer.  b) Calculate Ufinal / Uinitial , in terms of the parameters defined in the problem.

A capacitor stores potential energy PE when charged with a voltage of ΔV. If you double thevoltage, how much energy does the capacitor store?

Please asap

A 20 µF capacitor is connected to a 60 V battery until it becomes fully charged.  The capacitor is then disconnected from the battery and connected to a light bulb until 1/3rd of the charge is removed. Calculate the energy left in the capacitor.

I am not sure about the last parts d and e.

#16 Does the charge that flows into the capacitor during the charging go all the way through the capacitor and back to the battery, or does it get stored somewhere in the capacitor?  a It goes all the way through the capacitor and back to the battery.  b It gets stored somewhere in the capacitor.