If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance:

Assertion: When a battery remains connected,the electric potential energy increases if a dielectric material is inserted between the plates of a capacitor.
Reason: When a battery remains connected,the charge on the plates of the capacitor remains the same.

$A$ capacitor has some dielectric between its plates and the capacitor is connected to a $D.C.$ source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance,the energy stored in it,electric field,charge stored,and the voltage will increase,decrease,or remain constant.

Two identical capacitors $A$ and $B$ are connected in series to a battery of $E$.$M$.$F$. $E$. Capacitor $B$ contains a slab of dielectric constant $K$. $Q_A$ and $Q_B$ are the charges stored in $A$ and $B$. When the dielectric slab is removed,the corresponding charges are $Q_A^{\prime}$ and $Q_B^{\prime}$. Then:

$A$ 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:

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?