Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$, then the potential difference across the other capacitor will become:

If two charges $q_1$ and $q_2$ are separated by a distance $d$ and placed in a medium of dielectric constant $K$, what will be the equivalent distance between the charges in air for the same electrostatic force?

$A$ parallel plate capacitor has a plate separation of $0.01\, mm$ and uses a dielectric (whose dielectric strength is $19\, kV/mm$) as an insulator. The maximum potential difference that can be applied to the terminals of the capacitor is......$V$

Two point charges are kept in air with a separation $r$ between them. The force between them is $F_1$. If half of the space between the charges is filled with a dielectric of dielectric constant $K=4$,the force between them becomes $F_2$. If $1/3$ rd of the space between the charges is filled with a dielectric of dielectric constant $K=9$,then the ratio $F_1/F_2$ is:

$A$ slab of material with a dielectric constant of $3$ has the same area as the plates of a parallel plate capacitor but has a thickness of $\left(\frac{3}{4}\right) d$,where $d$ is the separation between the plates. What is the electrical potential difference between the plates when the slab is inserted? The initial electrical potential difference is $V_0$.

Which one statement is correct? $A$ parallel plate air condenser is connected to a battery. Its charge,potential,electric field,and energy are ${Q_0}$,${V_0}$,${E_0}$,and ${U_0}$ respectively. $A$ dielectric slab is inserted to fill the complete space between the plates while the battery remains connected. Now,the corresponding values $Q$,$V$,$E$,and $U$ are related to the initial values as: