Question

The relative permittivity of water is $81$ . If ${\varepsilon _0}$ and ${\varepsilon _w}$ are the permittivities of vacuum and water respectively, then:
(A) ${\varepsilon _0} = 9{\varepsilon _w}$
(B) ${\varepsilon _0} = 81{\varepsilon _w}$
(C) ${\varepsilon _w} = 9{\varepsilon _0}$
(D) ${\varepsilon _w} = 81{\varepsilon _0}$

Answer 1

Hint : To solve this question, we need to use the basic definition of the relative permittivity of a medium. Then, using the values given in the question, we will get the required relation.

Formula Used: In this solution we will be using the following formula,
$\Rightarrow {\varepsilon _r} = \dfrac{\varepsilon }{{{\varepsilon _0}}}$ where ${\varepsilon _r}$ is the relative permittivity, $\varepsilon$ is the permittivity of the medium and ${\varepsilon _0}$ is the permittivity of vacuum.

Complete step by step answer
The permittivity of a medium in electrostatics is defined with respect to its response to the electric field which is applied in the medium. The medium which gets more polarized, and therefore stores more electrical energy for a given electric field, is said to have higher permittivity.
The electric permittivities of all the mediums are expressed with the vacuum as a reference. So, instead of expressing their absolute permittivity values, their relative permittivity is expressed. The relative permittivity of a medium is defined as the ratio of the absolute permittivity of that medium to the permittivity of the vacuum. So, the relative permittivity is given as
$\Rightarrow {\varepsilon _r} = \dfrac{\varepsilon }{{{\varepsilon _0}}}$
Now, according to the question, the relative permittivity of water is $81$ , and the absolute permittivity of water is ${\varepsilon _w}$ . Substituting these above, we get
$\Rightarrow 81 = \dfrac{{{\varepsilon _w}}}{{{\varepsilon _0}}}$
Multiplying by ${\varepsilon _0}$ both the sides, we get
$\Rightarrow {\varepsilon _w} = 81{\varepsilon _0}$ .

Note
The relative permittivity of a medium is more commonly known as its dielectric constant. It is used in many engineering applications. While designing a capacitor, its dielectric constant is taken into consideration. These are also used to measure environment factors such as temperature, humidity etc. Since the dielectric constant varies with these factors, so the sensors constructed to detect the change in the dielectric constant can be used for this purpose.