Putting a dielectric substance between two plates of condenser, capacity, potential and potential energy respectively-

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be:

A parallel-plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants$k_1$, $k_2$, $k_3$ and $k_4$ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by

A simple pendulum of mass ' $m$ ', length ' $l$ ' and charge $\text{'}+q^{\text{'}}$ suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be

If ${\mathrm{q}}_{\mathrm{f}}$ is the free charge on the capacitor plates and ${\mathrm{q}}_{\mathrm{b}}$ is the bound charge on the dielectric slab of dielectric constant $\mathrm{k}$ placed between the capacitor plates, then bound charge ${\mathrm{q}}_{\mathrm{b}}$ can be expressed as :

For a capacitor, the distance between two plates is $5 \mathrm{x}$, the electric field between them is ${\mathrm{E}}_0$, now dielectric slab having dielectric constant $3$ and thickness $\mathrm{x}$ is placed between them in contact with one plate. In this condition what is the potential difference between its two plates is: