$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $K = 5$ as shown in the figure,the percentage increase in the capacitance will be......$\%$

$A$ parallel plate air capacitor is charged up to $100 \,V$. $A$ plate $2 \,mm$ thick is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by $1.6 \,mm$. The dielectric constant of the thick plate is

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?

$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

The distance between the plates of a charged parallel plate capacitor is $5 \ cm$ and the electric field inside the plates is $200 \ V/cm$. An uncharged metal bar of width $2 \ cm$ is fully immersed into the capacitor. The length of the metal bar is the same as that of the plate of the capacitor. The voltage across the capacitor after the immersion of the bar is......$V$

Two identical capacitors $1$ and $2$ are connected in series. Capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0 \text{ volts}$. The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charges stored in the capacitors before removing the slab,and $Q'_1$ and $Q'_2$ be the values after removing the slab. Then: