Hi Ajreen Qammar ,

could you please attach Fig. 1d, too? Where is the text taken from?

please check the attached paper

Hi Ajreen Qammar ,

please have a look at table II and figure 5 in the article behind the first link given below.

Table II gives a thickness of 7.4*10-10 m of the Stern layer for NaCl, and a capacitance density of 0.51 F/m2. These values are for 0.05 M NaCl, while the paper you attached deals with 0.1 M NaCl. Since the capacitance is proportional to the squareroot of the ion concentration, the capacitance density for 0.1 M is about sqrt(2) * 0.51 F/m2 = 0.72 F/m2.

Article Determination of Surface Potential and Electrical Double-Lay...

The Thomas-Fermi layer forms a second capacitance, in series with the Stern layer capacitance. The Thomas-Fermi screening length in gold is about 50 pm (see the paragraph above eq. (2) in):

Article Quantitative Cross-Platform Performance Comparison between D...

In the article you attached it says: "a few Angstroem"; I guess this difference is due to the fact that we are dealing with an exp-function; while 50 pm is the length constant, the value of the exp-function is less than 10-2 (i.e. it vanishes for practical purposes) at about 5 constants = 250 pm.

The relative permittivity of gold can be assumed to be 5 (sorry, can't find the source at the moment), so this value together with a thickness of 50 pm gives a capacitance density of 0.885 F/m2.

Both capacitances in series result in a total capacitance density of 0.397 F/m2.

On page 74 in the following book, curve (c) in Fig. II-14 gives the capacitance density as a function of the electrode potential. Instead of NaCl NaF is used, and instead of a gold electrode it's mercury. However, since the concentration of Na ions is the dominant parameter, both cases should give similar results; here the capacitance density at V = 0 is a bit more than 0.2 F/m2.

https://books.google.de/books?id=txfSBwAAQBAJ&pg=PA72&lpg=PA72&dq=capacitance+thomas-fermi+layer+gold&source=bl&ots=c4u-78M8M-&sig=ACfU3U3p6r2f2ZFYaAtzpGZhqV--fAWLJg&hl=de&sa=X&ved=2ahUKEwjHpoe_j9TmAhWSunEKHVzECxIQ6AEwAHoECAgQAQ#v=onepage&q=capacitance%20thomas-fermi%20layer%20gold&f=false

So, the truth is somewhere in the range 0.2 to 0.4 F/m2.

The figure shows that the dependence on the potential is not symmetric to V = 0.

With a sensing length of 10 mm, and a core diameter of 0.4 mm the capacitance lies in the range 2.5 to 5 uF.

Sorry, I just realized that you asked how to calculate these values, and I simply threw a few numbers at you; but I hope that the references I included will answer your question at least partly. We have the equation

C = epsilon0 * epsilon_r * A / d

but the problem here is that there is a sharp boundary (gold / electrolyte) on one side but the second boundary (of both the inner and the outer capacitor) is defined by an exp-function instead of a Heaviside step function; so d is a "soft value". Additionally, if there were an electrically insulating material between the conductors, one could just look up the associated epsilon_r in a table; but here the task is to determine a kind of "effective" epsilon_r.