Question

If a NaCl crystal is subjected to an electric field of 1000V/m and resulting
polarization is 4.3x10^-8 c/m2, dielectric constant of NaCl is​

Answer 1

Given info : If a NaCl crystal is subjected to an electric field of 1000V/m and resulting

polarization is 4.3 x 10^-8 C/m²

To find : dielectric constant of NaCl crystal is..

solution : dielectric constant is nothing but relative permittivity of medium.

we know, relation between relative permittivity, electric field and polarization is given as

$\quad P=\epsilon_0(\epsilon_r-1)E$

⇒$\epsilon_r=1+\frac{P}{\epsilon_0 E}$

here P = polarization = 4.3 × 10^-8 C/m²

ε₀ = permittivity in vacuum = 8.854 × 10^-12 C²/(Nm²)

E = electric field = 1000 V/m

now, $\epsilon_r$ = 1 + (4.3 × 10^-8)/(1000 × 8.854 × 10^-12)

= 1 + (4.3 × 10)/(8.854)

= 1 + 4.856

= 5.856

Therefore the dielectric constant of NaCl crystal is 5.856

Answer 2

Answer:

Explanation: If an ionic crystal is subjected to an electric field of 1000 V.m–1 and the

resulting polarization 4.3 × 10–8 m2. Calculate the dielectric constant of NaCl.

Given that, ε0 = 8.854 × 10–12 F.m–1.