Why is the derivative of the potential at a surface assumed to be continuous even though there is surface charge?

[per persson]

Homework Statement

Why is the derivative assumed to be continuous?

Relevant Equations

dV_2/dn-dV_1/dn=sigma/epsilon

I don't understand how why the derivative is continuous. Clearly there is a surface charge and by gauss law we should have

 

[kuruman]

Where did the equations that you posted come from?
Are they supposed to be the solution to the problem you posted or some other problem?

I am particularly suspicious of this

What is $\rho$ in the region $r\leq a$ when there is vacuum in that region? Answer: Zero.

In any case, if the inner shell is held at potential $V_0$, in the region $r\leq a$ the potential is also $V_0$. The derivative "just inside" must be zero while equal to $\dfrac{\sigma}{\epsilon_0}$ "just outside." Thus the derivative is discontinuous across $r=a$, the discontinuity being proportional to the surface charge density.

I see two possibilities

1. The equations that you posted are correct but are meant to be solutions to some other problem.
2. The equations that you posted are meant to be solutions for the problem that you posted in which case they are incorrect.