The lower half region of parallel plate capacitor is filled by a dielectric slab as shown in figure. Taking plate area as $A$ then incorrect statement is :-

         

 

What is a capacitor? You know that the capacity of a parallel plate condenser is $C=\frac{{\epsilon }_0 A}{d}$. What will be its new capacitance $C\text{'}$ of the capacitor if a material of dielectric $K$ is inserted between the plates? Calculate the capacitance of the capacitor shown below.

The capacity of a parallel plate capacitor with air is $18 \mathrm{pF}$ When a dielectric material is inserted in the space between the plates, its capacity becomes $108 \mathrm{pF}$.Calculate the permittivity of the material. What is the material?(${\mathrm{\epsilon }}_0=8.854\times {10}^{-12}{\mathrm{C}}^2{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}$)

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 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:

Explain the behaviour of dielectrics in an external field.