Electronic – Diode breakdown

How can a charge neutral substance have a potential?

To say that a PN junction has built-in potential isn't to say that the PN junction has a potential relative to ground or infinity etc.

How can there be an inherent potential in a doped semi-conductor if it is charge neutral?

Charge has been separated within the PN junction and, thus, there is an electric field across the depletion region and an associated potential difference. A charged capacitor is neutral but there is a potential difference (voltage) across the dielectric. There are some similarities but...

However, I don't understand why we can't measure the voltage drop across the PN block with a voltmeter.

As explained, for example, here, the built-in potential is not readily measured with, e.g., a voltmeter. In other words, this question has been asked here several times (which means you are not the only one perplexed by this - most are at first) and there are good answers already available.

1) The DC "resistance" is simply \$V/I\$, which in your case is \$0.7\cdot2k / (10 - 0.7) = 150\Omega\$. The dynamic resistance is \$\frac{dV}{dI}\$ of the diode, which can be easily found on the diode datasheet. Like this, for example:

_{(source: davidbridgen.com)}

Here \$\frac{dV}{dI}\$ is the inverse of the slope of the line around the \$0.7V\$ point. As you can see the line is almost vertical, so the slope is very big. So the inverse is very small.

2) The voltage across the diode is not \$0.01V\$. It is \$0.7\pm 0.01V\$. But since the dynamic resistance is the derivative, the DC component is going away.