The radii of two spherical conductors $A$ and $B$ are in the ratio $3:5$ Conductor $\text{'}A\text{'}$ is in air while $B$ is surrounded by a meditm of relative permittivity $6$. The ratio of the capacitances of $A$ and $B$ is

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:

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 :

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

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