Suppose you have 60Hz at 1 ampere (377 amps/second), 1 meter away from your PCB. Assume your PCB as 1 cm^2 loop at the input. How much interference will you pick up?
V = 2e-7 * Area/Distance * dI/dT
= 2e-7 * (1cm * 1cm)/ 1 meter * 377
= 2e-7 * 1e-4 * 377
= 1e-11 * 800
= about 8 nanoVolts.
If the distance is closer, you get more trash.
If the PCB area, or the differential-wiring area, is greater, you get more trash.
If the 60Hz has current surges from rectifier diodes (usually near the sin wave peaks), you get more trash.
If the current is from a black brick switch-reg, the dI/dT easily is 1,000X faster, and you get more trash.
What are the degrees-of-freedom, to reduce Vinduce and/or reduce risks?
ensure all power cords have the hot and return wires very close (as power cords are made) and have the hot/return wires be twisted-pairs
have input-surge-filtering on all 60Hz power supplies, so the rectifier diode surges are not 10amps/10 microseconds but more like 10 amps / 1 millisecond.
have steel shielding (thin galvanized steel) around all switching supplies
lay out your PCB with +/- traces only 10mils apart
now about those skin-sensor conductive pads and the wiring: enormously vulnerable to magnetic fields and to electric fields---- its a tough world.
Notice the computed level is 8 nV @ 60 Hz. For a setup that has the HOT power wire at infinite distance from the RETURN power wire. Normal power cords have the HOT next to (2mm apart) the RETURN wire, thus expect another 10:1 or 100:1 reduction. This is why high-end-audiophiles use special power cords, when their power-amplifiers draw 100 amp peaks, with ugly fast diode-turnon current-surges.
By the way, the initial formula comes from
Vinduce = [MUo * MUr * Area / (2 * pi * Distance)] / dI/dT
in this topology
simulate this circuit – Schematic created using CircuitLab