Some questions on the photoelectric effect

I was reading about the photoelectric effect and a few questions come to my mind.

For the context, I read that if you shine some light (of high energy) on a metal, this latter will emit electrons.

So I was wondering:

• how can a metal lose an electron? Would electrons 'fly' in the air? Or are we just talking about the metal having more "free" electrons?
• if electrons are freed 'in the air', would that mean the metal would become electropositive? I recall in my chemistry classes at high-school when making batteries, we needed a salt bridge between the oxydation and reduction half cells to maintain their electrical neutrality. Do metals need electrical neutrality?
• again if the metal "loses" electrons in the air: if we radiate enough light would the metal end up being short of electrons? as in, would the photo electric effect stop working on a given piece of metal after e.g an hour?

(note: I have read many questions on this website, including Photoelectric effect vs. electronegativity? and the wikipedia pages)

• "Without the current of electrons flowing from anode to cathode, a cloud of electrons will established around the cathode" ─ not quite. Electrons will fly away, but as they do so, the cathode will start to develop a positive charge, and, depending on the geometry (specifically, the cathode's self capacitance), this will produce an additional voltage $V_{\rm c\infty}$ compared to infinity, which acts in the same way as the bias voltage used in normal experiments. (cont.) Commented Nov 11, 2019 at 11:06
• The kinetic energy gained by electrons at infinity will then be $K = h\nu - \Phi - V_{rm c\infty}$, with $\Phi$ the work function of the metal. Once $V_{rm c\infty}$ becomes greater than $\Phi-h\nu$, electrons can no longer be emitted by the cathode: they will attempt to fly away, but they'll be trapped in the charged cathode's Coulombic electrostatic potential well, and they'll fall back to the cathode. Commented Nov 11, 2019 at 11:08