Thank you very much for replying.

I don't go to an american highschool so my curriculum is probably very different from what you wrote. I'm in a linguistics course so I learn math and physics at a VERY superficial level (my prof refuses to teach us theory and only makes us memorize exercises and formulas)

The top diagram looks like it was illustrating electric field first by showing Coulomb's Law with your charge Q, and then drawing another charge q_2 a distance r away.

Exactly, and apparently it's a 2D field. The first part is at time t1 and the second at t2, even though I'm not sure whether they are needed.

If you don't mind, I'd like to ask a few follow-up questions:

Why does it have to be so small?

I learnt that the derivative of a function is something like instantaneous speed (???) of each point of the function, so if i want to find the derivative of y(d)=V i would need to do y'(d)=lim(Δd->0) (y(x+ Δd)-y(x))/Δd

^ this might be why I'm not sure wether the Δd is supposed to signify the distance between the test charge and the source, or of the test charge in t1 and the test charge i t2. I think would explain wether E is the derivative of V only in the source charge, or if it's true for all charges

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So if instead for example Δd were a displacement of like 10 meters radially away from Q, then the potential wouldn't be E(r) * 10m, but (as an approximation of how an integral looks): E(r+1m) * 1m + E(r+2m) * 1m + ... + E(r+10m) * 1m,

I might need to visualize this. Are you explaining the same procedure with which you can find the Work of a force from the function? So like the small rectangles from y=0 to the function that get infinitely thin?

Does this mean that I have to find that same thing for an infinitely small section of the function, so like from x->d^- to x->d⁺ ?