$A$ parallel plate capacitor of capacitance $200 \,\mu F$ is connected to a battery of $200 \,V$. $A$ dielectric slab of dielectric constant $2$ is now inserted into the space between the plates of the capacitor while the battery remains connected. The change in the electrostatic energy in the capacitor will be ...... $J$.

$A$ parallel plate capacitor is connected to a battery. The quantities charge,voltage,electric field,and energy associated with the capacitor are given by $Q_0, V_0, E_0$,and $U_0$ respectively. $A$ dielectric slab is introduced between the plates of the capacitor,but the battery remains connected. The corresponding quantities now given by $Q, V, E$,and $U$ related to the previous ones are:

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 has a capacitance of $10 \ \mu F$ with air between its plates. Now,half of the space between the two plates is filled with a dielectric material of dielectric constant $K = 4$ as shown in the figure. Find the capacitance of the capacitor in $\mu F$.

Two parallel plate air capacitors of same capacity $C$ are connected in parallel to a battery of e.m.f. $E$. Then,one of the capacitors is completely filled with a dielectric material of constant $K$. The change in the effective capacity of the parallel combination is:

$A$ capacitor of capacitance $2 \mu F$ is charged to $50 V$ and then disconnected from the source. Later,the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by $25 \%$ of its initial value,then the dielectric constant of the dielectric material is