An external electric field of strength $E = 1.0 ext{ kV/cm}$ produces polarization in water. Assum

Question

An external electric field of strength $E = 1.0 	ext{ kV/cm}$ produces polarization in water. Assume this macroscopic polarization is equivalent to the perfect alignment of only one out of every $N$ water molecules (while the rest are randomly oriented). Find $N$ approximately. (Given: electric dipole moment of a water molecule $p = 0.62 	imes 10^{-29} 	ext{ C}\cdot	ext{m}$, dielectric constant of water $\epsilon_r = 81$, density of water $ho = 1000 	ext{ kg/m}^3$, Molar mass $= 18 	ext{ g/mol}$, $N_A = 6.02 	imes 10^{23} 	ext{ mol}^{-1}$, $\epsilon_0 = 8.85 	imes 10^{-12} 	ext{ F/m}$)

Professor PC Thomas Classes · Answer

The number density of water molecules is $n = \frac{ho N_A}{M} = \frac{1000 	imes 6.02 	imes 10^{23}}{0.018} \approx 3.34 	imes 10^{28} 	ext{ m}^{-3}$. If all molecules were perfectly aligned, the maximum possible polarization would be $P_0 = n p = (3.34 	imes 10^{28}) 	imes (0.62 	imes 10^{-29}) \approx 0.207 	ext{ C/m}^2$. The actual macroscopic polarization produced by the field is $P = \chi_e \epsilon_0 E = (\epsilon_r - 1)\epsilon_0 E$. Substituting the values ($E = 10^5 	ext{ V/m}$): $P = (81 - 1) 	imes (8.85 	imes 10^{-12}) 	imes 10^5 = 80 	imes 8.85 	imes 10^{-7} \approx 7.08 	imes 10^{-5} 	ext{ C/m}^2$. The problem states that the actual polarization is equivalent to $1$ out of $N$ molecules aligned perfectly, so $P = \frac{P_0}{N} \implies N = \frac{P_0}{P}$. Therefore, $N = \frac{0.207}{7.08 	imes 10^{-5}} \approx 2928$, which is closest to $2930$.