Question

An isotropic material of relative permittivity is placed normal to a uniform external electric field with an electric displacement vector of magnitude 6 * 10-4 C/m2 . If the volume of the slab is 0.7 m3 and magnitude of polarization is 5 *10-4 C/m2 . Find the value of PERMITTIVITY and total dipole moment of the slab.
​

Answer 1

Answer:

The value of permittivity is $53\times 10^{-12}F/m$ and dipole moment is $3.5 \times10^{-4}Cm$.

Explanation:

Given magnitude of electric displacement vector, of  $D=6 \times 10^{-4} C/m^{2}$

Magnitude of polarization,  is $P=5\times 10^{-4} C/m^{2}$

Volume of the slab, $V= 0.7 m^{3}$

Electric permittivity of free space, $\epsilon_{0} =8.85\times 10^{-12} F/m$

Polarization, $P=\epsilon_{0} (\epsilon_{r}-1)E=\epsilon_{0} \epsilon_{r}E-\epsilon_{0}E=D-\epsilon_{0}E$

$\implies P=D-\epsilon_{0}E\implies 5 \times 10^{-4} =6 \times 10^{-4} -8.85\times 10^{-12} \times E$

$\implies E=\frac{1}{8.85\times 10^{-8}} =11.3\times 10^{6}V/m$

From the constituent relation, $D=\epsilon E\implies \epsilon=\frac{D}{E}$    

$\epsilon=\frac{6\times 10^{-4}}{11.3\times 10^{6}}=0.53\times 10^{-10}=53\times 10^{-12}F/m$

Therefore, the value of permittivity is $53\times 10^{-12}F/m$

Value of relative permittivity, $\epsilon_{r}=\frac{\epsilon }{\epsilon_{0}} =\frac{53\times 10^{-12}}{8.85\times 10^{-12}}=5.98$

Polarization(P) is also defined as induced dipole moment(p) per unit volume,

$P=\frac{p}{V}\implies p=P\times V=5 \times10^{-4} \times0.7 =3.5 \times10^{-4}Cm$

Therefore, the dipole moment is $3.5 \times10^{-4}Cm$