
Dielectric constant of water is $80.$ What is its permittivity?
Answer
431.9k+ views
Hint:The permittivity can be defined as the opposition offered by any material and an ability of the material to store the electrical potential energy under the influence of an electric field. It is denoted by epsilon ($\varepsilon $ ). It can be measured by using the formula.
Complete step by step answer:
Dielectric constant of water, $k = 80$
Also, we know that –
The permittivity of free space is the capability of a vacuum to permit the electric fields and mostly used in electromagnetism.
The permittivity of free space ${\varepsilon _0} = 8.85 \times {10^{ - 12}}Farad/meter$
According to the formula –
$k = \dfrac{\varepsilon }{{{\varepsilon _0}}}$
Do cross-multiplication and make the unknown permittivity $\varepsilon $ as the subject –
$\varepsilon = k{\varepsilon _0}$
Place the given values in the above equations and simplify
\[
\varepsilon = 80 \times 8.854 \times {10^{ - 12}} \\
\therefore\varepsilon = 708.32 \times {10^{ - 12}}F/m \\
\]
Therefore, the required answer is – When the dielectric constant of water is $80$, then its permittivity will be \[\varepsilon = 708.32 \times {10^{ - 12}}F/m\].
Note:Remember the difference between the permittivity and permeability. Permittivity is the measure of the obstruction produced by the material whereas; the permeability is the ability of the material to allow the magnetic lines. Always remember and check the units of the given terms. The SI unit of the permittivity is farad per meter, the MKS (Meter Kilogram System) its unit is coulomb square per Newton into metre square, (${C^2}/N.{m^2}$ ). All the quantities should have the same system of units.
Complete step by step answer:
Dielectric constant of water, $k = 80$
Also, we know that –
The permittivity of free space is the capability of a vacuum to permit the electric fields and mostly used in electromagnetism.
The permittivity of free space ${\varepsilon _0} = 8.85 \times {10^{ - 12}}Farad/meter$
According to the formula –
$k = \dfrac{\varepsilon }{{{\varepsilon _0}}}$
Do cross-multiplication and make the unknown permittivity $\varepsilon $ as the subject –
$\varepsilon = k{\varepsilon _0}$
Place the given values in the above equations and simplify
\[
\varepsilon = 80 \times 8.854 \times {10^{ - 12}} \\
\therefore\varepsilon = 708.32 \times {10^{ - 12}}F/m \\
\]
Therefore, the required answer is – When the dielectric constant of water is $80$, then its permittivity will be \[\varepsilon = 708.32 \times {10^{ - 12}}F/m\].
Note:Remember the difference between the permittivity and permeability. Permittivity is the measure of the obstruction produced by the material whereas; the permeability is the ability of the material to allow the magnetic lines. Always remember and check the units of the given terms. The SI unit of the permittivity is farad per meter, the MKS (Meter Kilogram System) its unit is coulomb square per Newton into metre square, (${C^2}/N.{m^2}$ ). All the quantities should have the same system of units.
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