Determine the current flowing through the circuit and the potential...

From the given circuit diagram, the positive terminal of the battery is connected to the n-side (cathode) of the diode through the resistor, and the negative terminal is connected to the p-side (anode). This indicates that the diode is reverse-biased.

In a reverse-biased p-n junction, the current flowing through the circuit is equal to the minority carrier drift current (reverse saturation current).

Given the drift current is $20\mu\text{A}$, the current in the circuit is:
$I = 20\mu\text{A} = 20 \times 10^{-6}\text{ A}$

The potential difference across the resistor $R$ is:
$V_R = I \times R = (20 \times 10^{-6}\text{ A}) \times 20\Omega = 400 \times 10^{-6}\text{ V} = 0.0004\text{ V}$

Using Kirchhoff's voltage law, the potential difference across the diode $V_D$ is the total voltage minus the voltage drop across the resistor:
$V_D = V - V_R = 5.0\text{ V} - 0.0004\text{ V} = 4.9996\text{ V}$