Which elimination reactions, E1 or E2, depend on the concentration of the substrate?

#"E2"#.

(An #"E1"# reaction only has one substrate, so its concentration does not dictate the possibility of the reaction occurring. If it's there, and if it's reactive enough, it reacts until it's all gone.)

With this, we need to differentiate between the reactant and what is being reacted with.

The substrate is what is being reacted with. There must be something reacting with it, since in order for the concentration of the substrate to matter, it has to matter with respect to another reactant. Therefore, it must be a bimolecular reaction---a reaction involving two molecules.

It is clear that a bimolecular reaction is one that depends on the concentration of the substrate (whether it's the substrate itself or its conjugate base), because if there are two dependent reactants, when one runs out, the reaction is over. (It is the basic type of reaction you learned in General Chemistry where you have a limiting reactant and an excess reactant.)

When comparing #"E1"# and #"E2"#, notice that #"E2"# literally means "elimination second order."

A second-order reaction is either second order with respect to one reactant or first order with respect to each of two reactants.

#r(t) = k[A]^2#

or

#r(t) = k[A][B]#

where #A# and #B# are the reactant and the substrate.

#\mathbf"E2"# is first order with respect to each of two reactants.

There is a faster scenario, which occurs in one step. It is first order with respect to the substrate. This is known as #"E2"#.

The slower scenario is also first order with respect to the substrate, but there is another step---a slower second step---and that is first order with respect to the conjugate base of the substrate. This only happens if the carbanion is stabilized and the leaving group is a poor one. This is known as the less-common #"E1cB"# ("elimination first order conjugate base"), which is basically a two-step #"E2"# in which the second step is the rate-determining step.

An example one-step #"E2"# reaction mechanism looks like this, where on the substrate, the proton is antiperiplanar to the leaving group #("Br")#:

The proton eliminated was on the #beta#-carbon, while the leaving group was on the #alpha#-carbon. Therefore, this might also be known as #beta#-elimination.