Question

If an ionic crystal is subjected to an electric field of 1000 Vm–1 and the resulting polarization 4.3 × 10–8 cm2. Calculate the relative permittivity of NaCl.​

Answer 1

Answer:

Eletric field

E = 1000 Vm

P

= 4.3 x 10-8 cm2

P = ɛ0(E, -1)E =

&r 1+

h&r = 1+

P

4.3 x 10-8

(8.85 x 10-12 x 100)

&r = 5.86

80 8.85 x 10-12 Fm-!

Explanation:

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Answer 2

Given: ionic crystal is subjected to an electric field of $1000 Vm^-1$

To find: the relative permittivity of NaCl?

Explanation:

as we have,

                      electric field (E) = $1000 Vm^-1$

                                    and  ($e_{0}$) = $8.85$×$10^-12 Fm^-1$

        Resulting polarization (P) = $4.3$×$10^-8 cm^2$

we know that the formula of resulting polarization,

                                              $P = e_{o}(e_{r} - 1)E$

                                               $e_{r} = 1+\frac{P}{e_{o}E}$

now we have the formula of relative permittivity. let's find the relative permittivity of NaCl solution by putting values,

                                                $e_{r}$ $=$ $1+$${($$4.3$×$10^-8$÷$8.85$×$10^-12$×$100$$)$

                                                $e_{r}$ $=$ $5.86$

hence the relative permittivity of NaCl is $5.86$