$A$ capacitor, when filled with a dielectric $K = 3$, has charge $Q_0$, voltage $V_0$, and electric field $E_0$. If the dielectric is replaced with another one having $K = 9$, the new values of charge, voltage, and field will be respectively:

The figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plates of both capacitors is filled with a dielectric of dielectric constant $K = 3$. What will be the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric?

$A$ parallel plate capacitor is constructed with two plates. Each plate has an area of $100 \ cm^2$ and they are separated by a distance of $1 \ mm$. $A$ dielectric with a dielectric constant of $5.0$ and a dielectric strength of $1.9 \times 10^7 \ V/m$ is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing dielectric breakdown.

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:

The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \,m$ and a fiber sheet of thickness $0.5 \times 10^{-3} \,m$. The dielectric constants of mica and fiber are $8$ and $2.5$ respectively. If the fiber breaks down at an electric field of $6.4 \times 10^6 \,V/m$, then the maximum voltage that can be applied to the capacitor is: (in $\,V$)

In the figure,a capacitor is filled with dielectrics. The resultant capacitance is