Dielectric Properties Questions and Answers

Answer: C

The correct statement is that barium titanate is ferroelectric but quartz is not.

Every ferroelectric is piezoelectric, but not every piezoelectric is ferroelectric. Barium titanate has a spontaneous electric polarisation below its Curie temperature of about 120 °C that can be reversed by an applied field, with a hysteresis loop, so it is ferroelectric.

Quartz lacks a centre of symmetry, so stress produces a polarisation and it is piezoelectric, but it has no spontaneous, switchable polarisation and so is not ferroelectric. That rules out the options calling quartz ferroelectric and the one denying it for barium titanate.

Answer: A

The power loss is proportional to \(\omega\).

For a dielectric in an alternating field, the power dissipated per unit volume is

\[ \begin{aligned} P &= \frac{1}{2}\omega\epsilon_{0}\epsilon''E_{0}^{2} \\ &= \frac{1}{2}\omega\epsilon_{0}\epsilon' \tan\delta\, E_{0}^{2} \end{aligned} \]

where \(\epsilon''\) is the loss factor and \(\tan\delta\) the loss tangent. Each cycle loses a fixed fraction of the stored energy, so for a given loss factor the power lost is proportional to the number of cycles per second, that is to \(\omega\). This is why dielectric heating (as in microwave ovens) works at high frequencies. The other powers of \(\omega\) do not follow from this relation.

Answer: A

The ratio is about 0.5.

By the Clausius-Mossotti relation, \(\frac{\epsilon - 1}{\epsilon + 2}\) is proportional to the total polarizability. At optical frequencies only the electronic polarizability can follow the field, and there \(\epsilon = n^{2}\):

\[ \begin{aligned} \frac{\alpha_{e}}{\alpha_{total}} &= \frac{n^{2} - 1}{n^{2} + 2} \div \frac{\epsilon_{s} - 1}{\epsilon_{s} + 2} \\ &= \frac{1.25/4.25}{4.6/7.6} = \frac{0.294}{0.605} \\ &\approx 0.49 \end{aligned} \]

So about half of the static polarizability of NaCl is electronic and the rest ionic. Using the plain ratio \((n^{2} - 1)/(\epsilon_{s} - 1) = 0.27\) instead of the Clausius-Mossotti form gives no option.

Answer: A

The ratio of ionic to total polarizability is about 0.5.

By the Clausius-Mossotti relation the total polarizability is proportional to \((\varepsilon_{s} - 1)/(\varepsilon_{s} + 2)\). At optical frequencies the ions cannot follow the field, so only the electronic part responds, with \(\varepsilon = n^{2} = 2.25\):

\[ \begin{aligned} \text{total} &\propto \frac{5.6 - 1}{5.6 + 2} = \frac{4.6}{7.6} = 0.605 \\ \text{electronic} &\propto \frac{2.25 - 1}{2.25 + 2} = \frac{1.25}{4.25} = 0.294 \end{aligned} \]

\[ \begin{aligned} \frac{\alpha_{ion}}{\alpha_{total}} &= \frac{0.605 - 0.294}{0.605} \\ &= 0.51 \end{aligned} \]

which is 0.5 to one figure. 0.7 to 0.9 would follow from comparing \(\varepsilon_{s}\) and \(n^{2}\) directly instead of using the Clausius-Mossotti factors.