$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

Which one statement is correct? $A$ parallel plate air condenser is connected to a battery. Its charge,potential,electric field,and energy are ${Q_0}$,${V_0}$,${E_0}$,and ${U_0}$ respectively. $A$ dielectric slab is inserted to fill the complete space between the plates while the battery remains connected. Now,the corresponding values $Q$,$V$,$E$,and $U$ are related to the initial values as:

$A$ capacitor is charged by a battery and the energy stored is $U$. Now,the battery is removed and the distance between the plates is increased to four times its original value. The new energy stored becomes:

$A$ parallel plate capacitor is charged and then isolated. If the separation between the plates is increased,which one of the following statements is $NOT$ correct?

Half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,then the new capacitance will be: