Is it specific capacitance (Csp) is proportional to current density?

[N.F. Zain]

I've run the experiment. By varying the current and fix the mass of the electrode, I found out that the specific capacitance (Csp) is increased when applied current (I) is increased.

By using this formula,

Csp = 4I / [M*(dV/dt)],

where I is the applied current, M is the mass of both electrodes, and dV/dt is the discharging slope after the IR drop, respectively.

Current density = I/M,
where I is the applied current, M is the mass of both electrodes.

Logically speaking, we can see from the formula above that Csp is proportional to I, M is fixed for all samples, but most of the published journals have reported that Csp is decreased when current density (I/M) is increased.

How could this happen? Please somebody help me with the brief explanation. TQVM. :(

Here I attached the related journals to this problem,
1) http://iopscience.iop.org/0957-4484...&spage=13574&stitle=Proc.+Natl+Acad.+Sci.+USA
2)http://www.sciencedirect.com/science/article/pii/S0013468606010851

 

[Nate810]

The reason that the specific capacitance (Csp) is decreased when current density (I/M) is increased is because of the effect of the IR drop caused by the applied current. The IR drop is a voltage drop across the electrodes due to the resistance of the electrolyte, and it causes a decrease in the voltage between the two electrodes. As a result, the discharging slope of the capacitor (dV/dt) is decreased, leading to a decrease in the specific capacitance (Csp).