Question

If the dielectric constant of a medium is unity, what would be its relative permittivity?

Answer 1

We can start by defining what each of these terms mean. For instance, dielectric constant is a measure of the amount of electric potential energy, in the form of induced polarization that is stored in a given volume of material under the action of an electric field. Relative permittivity of a material is the ratio of capacitance of a capacitor while using that material as its dielectric to the value of capacitance of the capacitor while using vacuum as the dielectric.

Formulas used:
The dielectric constant of a material,
$\kappa = \dfrac{{{{\rm E}_m}}}{{{{\rm E}_0}}}$
Where, ${{\rm E}_m}$ (pronounced as epsilon m) is the permittivity of the material
${{\rm E}_0}$ is the permittivity of vacuum

Complete step by step answer:
The following data is given to us, the dielectric constant of the medium is, ${{\rm E}_m} = 1$.Relative permittivity is the same as dielectric constant as you can see from the hint part.
Therefore, the value of relative permittivity can be found out using,
$\kappa = \dfrac{{{{\rm E}_m}}}{{{{\rm E}_0}}}$
The value of permittivity of vacuum is a known constant, its value is
${{\rm E}_0} = 8.85 \times {10^{ - 12}}{m^{ - 3}}k{g^{ - 1}}{s^4}{A^2}$
Substituting this value in the equation above, we have,
${{\rm E}_m} = \kappa \times {{\rm E}_0}$
The calculations are done as follows,

\Rightarrow {{\rm E}_m}= 1 \times 8.85 \times {10^{ - 12}} \\\ \therefore {{\rm E}_m}= 8.85 \times {10^{ - 12}}$$ **Thus, if the dielectric constant of a medium is unity, $$8.85 \times {10^{ - 12}}{C^2}/N{m^{ - 2}}$$ will be the relative permittivity of the material.** **Note:** Dielectrics play an important role in capacitance of capacitor. A capacitor is an electronic component designed to store electric charge. This is widely built by sandwiching a dielectric insulating plate in between the metal conducting plates. The dielectric property plays a major role in the functioning of a capacitor.The layer made up of dielectric material decides how effectively the capacitor can store the charge. Picking the right dielectric material is crucial.