$A$ capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant $K=2$ is inserted between the plates. The potential difference across its terminals becomes

In a parallel plate capacitor of capacitance $C$,a metal sheet is inserted between the plates,parallel to them. If the thickness of the sheet is half of the separation between the plates,the new capacitance will be:

$A$ 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 ...... $J$.

$A$ 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:

$A$ parallel plate air capacitor has capacitance $C_p$. It is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$. Now its capacity becomes $C_K$. The ratio $C_P$ to $C_K$ is

The capacity of an air condenser is $2.0 \, \mu F$. If a medium is placed between its plates,the capacity becomes $12 \, \mu F$. The dielectric constant of the medium will be: