Question

The dielectric constant of water is 80 what is its permittivity

Answer 1

ε/ε₀=k(dielectric constant)
here, 
        k=80;ε₀=8.854*10^-12
∴ε = 80 * 8.854 * 10^-12 = 708.32 *10^-12 is the permittivity of water

Answer 2

The permittivity will be $708 \times 10^{-12} \ \mathrm{F} / \mathrm{m}$ when the water's dielectric constant (given) is 80

Given:

Dielectric constant of water (given) = 80

To find:

Permittivity $(\varepsilon)$ = ?

Solution:

$\text{Dielectric constant}=\frac{\text {Relative permittivity}}{\text {Permittivity of free space}}$

$\text{Relative permittivity}(\text {water})=\text {Dielectric constant} \times \text{Permittivity of free space}$

$\Rightarrow k=\frac{\varepsilon}{\varepsilon_{0}}$

Dielectric constant is termed as the ratio of relative permittivity to the free space permittivity and Relative permittivity is termed as the product of dielectric constant to the free space permittivity.

Where,

$\varepsilon = \text{ Relative permittivity}$

$\varepsilon_{0} = \text{Free space permittivity}$

We know that,

$\text {Permittivity of free space}=\varepsilon=8.85 \times 10^{-12} \ \mathrm{F} / \mathrm{m}$

$\Rightarrow \text { Relative permittivity(water) }=80 \times 8.85 \times 10^{-12}$

$\therefore \text { Permittivity of water }=\varepsilon=708 \times 10^{-12} \ F / m$