I'm thinking of the proofs for "charge accumulation at the points of a spiky conductor" e.g. https://www.feynmanlectures.caltech.edu/II_06.html 6-11
If charge accumulates at sharp edges, I'm wondering whether a spiky sphere stores more total charge than a smooth sphere at the same potential.
Yes, the large sphere at potential V would have held charge Q alone, and with a small sphere (spike) connected at potential V which would have held charge q alone, the entire structure should hold Q+q. It is not too surprising that adding more material adds to the capacitance, as it does in the case of parallel plate capacitors etc.
The source of diminishing returns is that the maximum voltage reachable is now lower. For example, you can increase capacitance of parallel plates by reducing the distance between them, d. $C \propto \frac{A}{d}$ for area A and distance d. But now the electric field $E=\frac{V}{d}$ will be larger, which risks overcoming the dielectric strength and discharging the charge (https://en.wikipedia.org/wiki/Dielectric_strength).
Similarly, the electric field of a spike tip will be large and will limit the maximum potential of the overall ball+spikes by an approximate ratio of (ball radius) / (spike radius of curvature). So if you have gained Q+q charge per potential, you have reduced your total possible potential and total possible charge by q/Q=(spike radius)/(ball radius). You can estimate the number of spikes you need to add to breakeven, $(1+N*q/Q)*(q/Q) = 1$, roughly $N = (Q/q)^2$. So if the tip of a spike is 1 mm for a 1 meter ball, you need a million spikes.
This behavior of spikes is one reason lightning rods are spike-shaped. They facilitate the breakdown of air and provide an easy conduit for charge differences between ground and air to equalize, even when lightning is not visible.
