The potential energy of a charged parallel plate capacitor is $U_0$. If a slab of dielectric constant $K$ is inserted between the plates,then the new potential energy will be

$A$ medium having dielectric constant $K > 1$ fills the space between the plates of a parallel plate capacitor. The plates have a large area,and the distance between them is $d$. The capacitor is connected to a battery of voltage $V$,as shown in Figure $(a)$. Now,the plates are moved such that the distance between them becomes $2d$,with the dielectric slab of thickness $d$ remaining in between,as shown in Figure $(b)$. In the process of going from the configuration depicted in Figure $(a)$ to that in Figure $(b)$,which of the following statement$(s)$ is(are) correct?

If a slab of insulating material $4 \times 10^{-3} \ m$ thick is introduced between the plates of a parallel plate capacitor,the separation between the plates has to be increased by $3.5 \times 10^{-3} \ m$ to restore the capacity to its original value. The dielectric constant of the material will be:

An air-filled parallel plate capacitor has capacity $C$. If the distance between the plates is doubled and it is immersed in a liquid,the capacity becomes twice. The dielectric constant of the liquid is:

$A$ parallel plate capacitor has a plate area of $100\, m^{2}$ and a plate separation of $10\, m$. The space between the plates is filled up to a thickness of $5\, m$ with a material of dielectric constant $10$. The resultant capacitance of the system is $'x'\, pF$. Given $\varepsilon_{0} = 8.85 \times 10^{-12} F \cdot m^{-1}$,the value of $'x'$ to the nearest integer is:

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