Question

Oil/water interfacial potential. Consider an uncharged oil/water interface. On the aqueous side, the proximity of the oil phase biases the orientation of the water molecules in an unknown way. Calculate the resulting potential across the boundary layer for the maximum bias, the complete line-up of the first layer of water dipoles perpendicular to the interface. Treat this layer as a parallel-plate capacitor, with one water molecule occupying \(10 \mathrm{~A}^{2}\) of the interfacial area. The dipole moment is \(\mu=1.85\) debye \(\left(1\right.\) debye \(\left.=3.336 \times 10^{-30} \mathrm{Cm}\right)\) per water molecule. Take two values of the dielectric constant \(D\) between the capacitor plates: (a) \(D=2\), as for oil, and (b) \(D=80\), as for bulk water. (If water is perfectly oriented-a situation called dielectric saturation-then \(D=2\) is more likely.)

Short answer

For D=2, V≈7.64V. For D=80, V≈0.19V.

Step-by-step solution

- Define the Concepts and Given Data

Identify the parameters provided: each water molecule has a dipole moment, \(\mu = 1.85 \, \text{Debye} = 1.85 \, \times 3.336 \, \times 10^{-30} \, \text{Cm}\). The area per water molecule at the interface, \(A = 10 \, \text{Å}^2 = 10 \, \times 10^{-20} \, \text{m}^2\). Dielectric constants: \(D = 2\) (oil side) and \(D = 80\) (bulk water side). Treat the first layer of water dipoles perpendicular to the interface like a parallel-plate capacitor.

- Understanding Parallel-Plate Capacitor Model

For a parallel-plate capacitor, the potential difference is given by the formula: \(V = \frac{Qd}{\epsilon_0 \epsilon_r A}\), where \(Q\) is the charge, \(d\) is the separation distance, \(\epsilon_0\) is the vacuum permittivity \((8.854 \, \times 10^{-12} \, \text{Fm}^{-1})\), and \(\epsilon_r\) is the dielectric constant. We need to find the charge on each plate in terms of the dipole moment aligned perpendicular to the plates.

- Calculate the Effective Charge per Dipole

The dipole moment \(\mu\) of water is related to effective charge \(q\) and separation distance \(d\), as \(\mu = qd\). Thus, \(q = \frac{\mu}{d}\). Since \(\mu = 1.85 \, \times 3.336 \, \times 10^{-30} \, \text{Cm}\), and each water molecule occupies an area of \(10 \, \times 10^{-20} \, \text{m}^2\), the thickness \(d\) of one water molecule layer is assumed to be approximately the diameter of a water molecule, about \(3 \, \times 10^{-10} \, \text{m}\).

- Calculate Potential Difference for Dielectric Constant D=2

Given \(D = 2\), inserting the values: \(q = \frac{1.85 \, \times 3.336 \, \times 10^{-30}}{3 \, \times 10^{-10}}\), and using \(\epsilon_0 = 8.854 \, \times 10^{-12} \, \text{Fm}^{-1}\), we calculate the potential difference: \(V = \frac{q d}{\epsilon_0 \epsilon_r A}\). After substituting the values, it can be shown that: \(V \approx 7.64 \, \text{V}\).

- Calculate Potential Difference for Dielectric Constant D=80

Similarly, for \(D = 80\), \(V = \frac{q d}{\epsilon_0 \epsilon_r A}\) substituting \(\epsilon_r = 80\), the potential difference can be calculated as: \(V \approx 0.19 \, \text{V}\).

Key concepts

Dielectric Constant

The dielectric constant, also known as relative permittivity, measures the material's ability to store electrical energy in an electric field. It's represented by \(\text{D or } \epsilon_r\). Higher dielectric constants mean better insulation properties and more stored electrical energy. Different materials have different dielectric constants. For example, oil often has a dielectric constant around 2, while water has a much higher value, approximately 80, which signifies that water can store significantly more energy compared to oil.
The dielectric constant is important in the context of the oil/water interface because it influences the potential difference across that boundary. Understanding and calculating with these constants allows us to predict how the materials will behave in electric fields.

Dipole Moment

The dipole moment is a measure of the separation of positive and negative electrical charges in a molecule. It's given by the product of the charge magnitude and the distance separating the charges, symbolized by \(\mu\). For a water molecule, the dipole moment is particularly important because water is polar, meaning it has a slight separation of charge leading to unique properties.
In our problem, each water molecule has a dipole moment of \(1.85\, \text{Debye} = 1.85 \times 3.336 \times 10^{-30}\, \text{Cm}\). When water molecules align at the oil/water interface, they create an effective charge distribution that influences the overall potential difference. This alignment can be considered in calculations as creating a layer with an equivalent effect to a parallel-plate capacitor.

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by an insulating material (dielectric). The potential difference across the plates is given by the formula: \( V = \frac{Qd}{\epsilon_0\epsilon_r A} \).
In this exercise, the oil/water interface is treated as a parallel-plate capacitor where water molecules, oriented perpendicular to the interface, act as dipoles. Each molecule's dipole moment can be broken down into an effective charge separation \( q = \frac{\mu}{d} \).
This helps in calculating the potential difference considering different dielectric constants (\(D = 2\) for oil and \(D = 80\) for bulk water). For \(D = 2\), the potential difference across the boundary, considering our given values, calculates approximately to \( 7.64 \text{V}\), while for \(D = 80\), it calculates to \( 0.19 \text{V}\). Understanding these calculations helps better predict and explain the electrostatic behavior at material interfaces.