If your voltage source is AC, then the changing electric fields, continually changing, allow a flow of current to the surrounding environmental conductors such as the floor if concrete or dirt, to the metal grids behind tile walls, and to the ever-present wall wiring of the 50/60Hz power system.
Lets model your feet, above concrete/grounded floor as 0.1meter * 0.1 meter, and assume the insulation is 1cm thick, with polarizability being 5X that of air or vacuum, thus 5X energy can be stored. How much capacity is this?
C = Eo * Er * Area/Distance
C = 8.98e-12 Farad/meter * 5 * 0.1m*0.1m / 0.01m
C ~~ 50pF * 0.01/0.01 = 50pF
What will be the (displacement, or field charging) current?
Q = C*V , now differentiate to get dQ/dT = dC/dT * V + C * dV/dT
and proclaim the C to be constant so dC/dT is zero. Result is
I = dQ/dT = C * dV/dT
For 117vac rms 60Hz, where the 60Hz is 377 radians/second, the slewrate the dV/dT is 117 * peak/rmsconvert * 377
dV/dT = 117 * 1.414 * 377 or approx. 70,000 volts per second. The current thru 50pF is
I = 50pF * 70,000 volt/sec = 5e-11 * 7e+4 = 35e-7 or 3.5 microAmperes at 60Hz.
On the other hand, if your voltage source is DC, once the electric-fields become stable, no more current/charge/electronics will move.