Yes, there are many neutral electrophiles.
One large class are the compounds with leaving groups. Some that you have likely encountered are: acids (HCl, HBr, sulfuric acid, etc) which are electrophilic at the hydrogen; unhindered alkyl halides/sulfonates (e.g., methyl iodide for SN2 reaction); dihalides (e.g., Br2 for electrophilic addition to an alkene).
A second large class are compounds with pi bonds to an electronegative atom. Most importantly this class includes the carbonyls (aldehydes, ketones, esters, etc). Michael acceptors fall into this class as well, which have an extended pi system containing an electronegative atom.
Since you are covering electrophilic aromatic substitution, soon you may learn about nucleophilic aromatic substitution, where a highly electron poor aromatic ring (e.g., having multiple nitro groups) bearing a leaving group can be attacked by a strong nucleophile.
In all of these cases, although the compounds are neutral, there is a partial positive charge at the electrophilic site explained by a polarized bond (acids, alkyl halides), resonance (carbonyl), or spontaneously (dihalides).
Neutral electrophiles are less reactive than their positively charged counterparts, which means that the nucleophiles are typically stronger for a reaction to occur.
As for negatively charged electrophiles, I can think of only a couple of cases that could be considered that way. For example, in the reduction of a carboxylic acid with lithium aluminum hydride, the first step is deprotonation of the acid to give a carboxylate. LiAlH4 is very reactive and adds a hydride to the carboxylate. That could be seen as nucleophilic attack on a negatively charged electrophile, and many people will draw it that way. In reality though, there is something stabilizing the negative charge, either the lithium or aluminum. So it's not really a negatively charged carboxylate that's being attacked. I don't think that there are really any negatively charged electrophiles, but someone may have an example I haven't seen or thought of.