The max current is not [typically] the capacitive current. Capacitive current is the background current in the CV, while peak currents come from Faradaic reactions. If you apply a linear sweep across a capacitor you get a constant current, with a slight rise time due to resistance slowing things down (RC time constant). There is, however, a catch: you must be using either a surface mode for data acquisition or a linear (analog) scan generator. Most (modern, commercial) digital potentiostats do not capture all of the capacitive current unless they have one of those two features.
Instead of (the current) i vs. Vr (scan rate) plot(s, in your question), you could plot (the calculated, nominal, Capacitance) C vs Vr. A CV (Cyclic Voltammetry) regime, having a pure Capacitive origin, converges to a limiting value CpureCap, a constant Capacitance value (CpureCap), in the C vs Vr diagram transformation.
So, you can consider[1] the ihigh (current) inside this Vr-regime, as a pure Capacitive current ( ihigh~ipureCap) for your target, e.g. calculate, only, the Capacitive (component of the) current density, from your curves' set of the exp. CV(s).
1. Name it, "i.Cap-condition"[2] for a Capacitive current.
2. In the case of a SC (SuperCapapacitor) you might find an additional (2nd converging) limiting value; the higher CpureCap value is due to the SC (condition).
The capacitance of a supercapacitor ((F/g) versus scanrate (mV/s) can be calculated in experimental session of followed report, and then you can see a resulted plot in Fig. 5f.
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