When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates,then the energy stored:

When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?

$A$ parallel-plate capacitor of plate area $10 \text{ cm}^2$ and plate separation $3 \text{ mm}$ is charged to a potential difference $12 \text{ V}$ and then the battery is disconnected. $A$ slab of dielectric constant $3$ is then inserted between the plates. The work done on the system in the process of inserting the slab is $\alpha \varepsilon_0$. The value of $\alpha$ is (Take $\varepsilon_0$ as the permittivity of free space).

$A$ parallel plate capacitor filled with a medium of dielectric constant $10$ is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$. Then the energy of the capacitor will ......................

$A$ parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates,then: