# Debye length

#### Prerequisites

In the previous reading on Electrical interactions in ionic solutions, we noted that the charge on a a DNA molecule in a salt solution experiences a screening effect as a result of the salt ions clustering near the DNA. As a result, the electric field produced by a charged biological macromolecule, or indeed by any unbalanced charges, should be different than it would produce in a vacuum.

What is happening is that the unbalanced charge draws up ions of the opposite sign that tend to cancel it. But the ions are in thermal motion and tend to move apart to increase entropy. The balance between these two effects produces a volume of ionic charge that effectively cancels the unbalanced charge, screening it from the rest of the fluid. Let's make this quantitative by using the physics we know to get an estimate of the distance scale.

To make this quantitative we would like to know two things:

- How large is the volume occupied by the positively charged ions that neutralize the negatively charged DNA — the “screening cloud”? Put differently, how far away from the DNA do we have to go before the electric field will drop essentially to zero?
- What is the mathematical function describing how the electric field and the electric potential depend on distance?

Let’s begin by reasoning about the physical parameters that determine the size of the screening cloud. Specifically, let’s model the screening cloud as extending a distance $λ_D$, known as the ** Debye length**, away from the DNA at every point. Inside the cloud there are many more positive ions than negative; outside the cloud there are equal concentrations of positive and negative ions. What determines this size $λ_D$?

- If temperature increases, the effects of entropy and temperature also increase. We'd expect the screening cloud to spread out due to the increased molecular motion. Therefore, $λ_D$ should increase as $T$ increases.
- If the
*undisturbed*concentration $c_0$ of salt ions increases, then the number of ions that have to be moved away from an even distribution in order to form the screening cloud is a smaller fraction of the total. Therefore, the*fractional change*in concentration required to screen the DNA is smaller. This means that the entropic cost of forming the screening cloud decreases as the salt concentration increases. Therefore, $λ_D$ should*decrease*as $c_0$ increases. - If the charge of the individual positive ions increases, fewer of them are needed to screen the DNA and again the entropic cost of forming the screening cloud is less. Therefore $λ_D$ should decrease as $ze$, the charge of one of the positive ions, increases.

The mathematical expression for the size of the screening cloud $λ_D$ is determined by solving differential equations to find the arrangement that balances electric potential energy and entropy. We won’t do that in this class. We’ll make use of the result, but we can get a good insight into the structure of the result by doing some estimations.

The central negative charge (blue charges in the small blue cloud) we are trying to screen drags in ions and they have extra PE because they are now closer (excess red charges in the pink sphere) than they would be if the charge weren't there.

The electrostatic potential energy for two charged particles of charge $ze$ separated by a distance $r$ is given by (using the dielectric constant to reduce the strength of the field due to simple polarization of molecules rather than the moving of ions)

$$U^{elec} = \frac{k_C z^2 e^2}{\kappa r}$$

The other effect that we have to consider is that the thermal energy of their motion tends to spread them out. Let's try estimating the size of our sphere by balancing the amount of PE against the amount of thermal energy.

Consider the ions in a volume of $(λ_D)^3$. Since the concentration is $c_0$, we have about $c_0(λ_D)^3$ ions in that volume. If total PE is about $k_CQ^2/λ_D$ and we balance that against $k_BT$ for each ion, we'll get something like

$$k_B T \sim \bigg(\frac{k_C Q^2}{\kappa \lambda_D}\bigg)(c_0\lambda^3_D) = \frac{1}{\kappa}k_Cz^2e^2 c_0\lambda^2_D $$

Solving for $λ_D$ gives

$$\lambda = \sqrt{{\frac{\kappa k_BT}{k_Cz^2e^2 c_0}}}$$

So the numerator inside the square root is the thermal energy $k_BT$, and the denominator of our expression for $λ_D$ is proportional to the electric potential energy with the length ($1/r$) left out.

A more mathematically rigorous treatment gives

$$\lambda = \sqrt{{\frac{\kappa k_BT}{8 \pi k_Cz^2e^2 c_0}}}$$

In addition, solving the differential equations gives the following form for the electric potential of a charged particle or sphere in salt solution:

$$V^{screened}(r) = k_C q \frac{e^{-r/\lambda_D}}{\kappa r}$$

So we see that the potential decreases exponentially compared to the potential of a charged particle in vacuum, and the distance over which the potential decays to $1/e$ of its value without screening is $λ_D$. The corresponding electric field is slightly more complicated (it’s the derivative of $V$ with respect to $r$) but is essentially the same in being exponentially screened compared to the electric field of a charged particle in vacuum:

$$E^{screened}(r) = k_C q \frac{e^{-r/\lambda_D}}{\kappa r}\bigg(\frac{1}{r} + \frac{1}{\lambda_D}\bigg)$$

The figure at the right shows a surface (a piece of a membrane) with some excess negative charge (shown in red) imbedded in an ionic fluid. If you count positive charges (shown in blue) in the fluid, you see that there are more of them near to the negative surface. They don't just come up to the negative charges and cancel them because they are jiggling due to thermal energy. The balance between those two effects produces a slightly increases density of positive ions, $c_+$ and a slightly decreased density of negative ions, $c_-$, as shown in the upper graph. The resulting screened potential, $V^{sc}$, falls to zero exponentially — much faster than a $1/r$ unscreened power law, $V^{unsc}$.

Catherine Crouch, Swarthmore College 2/12/12, figures by Joe Redish 5/13/19

Last Modified: May 24, 2019