It’s not even depend on computer, the only factor is whether the case is metal or plastic. You can measure the output of the cable to the ground without connecting it to a computer as the cable itself is metal.
This is not sustained 60+V shock. the 60V is “caused” by the capacitance of the circuit(as shown C1 below) and it’s “attenuated” by the capacitance to the Earth(as shown below as C2), so it won’t shock you like a touching a 60VAC live as the current is very low even when shorted. The leakage current is due to non-ideal circuit. The power supply has galvanic isolation so mains voltage can’t pass through. The resistance between socket and USB-C is infinite. However a non-ideal power supply and cable have tiny capacitance in it, and AC power can transfer across a capacitor. The real capacitance of the cable is very low so the leakage current is also very low(<10mA) so the leakage protection won’t trip.
![capacitor-coaxial-cable](https://static-community.frame.work/original/3X/2/0/201c1a11ab267a8a92d30f8ec5a734112d90e031.gif)
![nonidealcable](https://static-community.frame.work/original/3X/2/b/2bf3294470a93dce78b1b97f7fcece2adcf9dd1a.jpeg)
For a floating power supply
The output positive to negative is 20V but the output to ground voltage is undefined due to lack of reference voltage. In real-world application there’s tiny capacitance than cannot be ignored
As shown the capacitance C1(the power brick and the cable) C2(device to Earth) so the voltage is somewhat between 0 and mains as AC can pass through a capacitor. To experience the classic “MacBook electric shock/tingling/vibration” touch the output, i.e. between C1 and C2, you as a resistor from output to Earth as tiny current pass through your body. A grounded has a “cable” instead of C2, pulling the output ground, to Earth’s voltage i.e 0V and current won’t pass through you as it goes to the “shorter” route.
For example, if C2 is four times of C1, the “floating” voltage is 0.2 of the mains voltage.
As shown in my example of 72V, since the mains voltage is 230V, the C2 is about 2.19 times of C1. However since the multimeter is not ideal, it means a high resistance resistor in parallel of C2, the actual floating voltage is higher than measurement.
To measure the leakage current, C2 is shorted by a current meter. In my example above the current is 0.45mA, the mains voltage is 230V, the frequency is 50Hz, we can calculate the capacitance of the non-ideal SMPS and USB-C cable.
0.45mA = 230V * sqrt(2) * 2 * pi * 50Hz * C1
Thus, C1 = 4.4nF