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:

Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now be

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:

Two capacitors $C_1$ and $C_2$ are connected to a battery as shown in the figure. The space between the plates of $C_1$ is filled with air,and the space between the plates of $C_2$ is filled with a dielectric material. Which of the following is true regarding the charges $Q_1$ and $Q_2$ on the capacitors?

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.

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: