In the photoelectric effect, the current increases as the intensity of the light increases. Is there an equation for this? I've concluded the intensity of the light (which is proportional to $\frac{1}{r^2}$) is proportional to the change in flow of electrons per unit of time. Is there an explicit equation showing this relationship?
 A: Essentially, calculate the number of photoelectrons emitted per unit time. This likely requires calculating the incident flux. 
Note, a decent explanation from askiitians:

The number of photons incident on the plate per time (called photon
  flux) is given by
$$ \Phi = P' / E = (P/4 \pi r^2) A / hf \, .$$
If $f > f_0$ (threshold frequency) and photon efficiency of the metal
  plate is $\eta$, then the number of photoelectrons emitted per time is
  given by
$$n = \Phi \eta (P/ 4 \pi r^2) A / hf) \, .$$
Finally, the photocurrent i is given by
$$i = ne$$
where $e$ is the charge of an electron ($e = 1.6 × 10^{–19}$ Joules).

A: Once the photons have exceeded the energy threshold and are able to eject the electrons, you're correct that the intensity and electron ejection are proportional. I would expect that the proportionality constant is experimentally determined in each case. I've never seen an equation for it. 
A: Basically, the number of photoelectrons emitted is proportional to the number of photons. We assume/know that each photon is associated with removing an electron from the metal. As the number of photons increase(i.e the intensity increases) the number of electrons increases hence the current increases since I=ne. 
We can see this intuitively from the particle nature very easily.
