$A$ 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:

$A$ capacitor $C$ is charged to a potential difference $V$ and the battery is disconnected. Now,if the capacitor plates are brought closer slowly by some distance,then:

An air capacitor has a capacitance of $1 \mu F$. Now,the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is ($d=$ distance between two plates,$K_1=8$ and $K_2=4$ are the dielectric constants of the two dielectrics respectively).

$A$ parallel plate air capacitor is charged and then isolated. When a dielectric material is inserted between the plates of the capacitor,which of the following does not change?

When a dielectric material is introduced between the plates of a charged condenser,what happens to the electric field between the plates?

$A$ 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?