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

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The discussion centers on the continuity of the derivative of potential at a surface with surface charge, questioning how it can be continuous despite the presence of charge. It highlights that within a vacuum region, the charge density is zero, leading to a potential of V0 inside the inner shell. The derivative of the potential just inside the surface is zero, while just outside it equals σ/ε0, indicating a discontinuity at the surface. This raises doubts about the validity of the equations presented, suggesting they may either address a different problem or be incorrect for the stated scenario. The conversation emphasizes the need for clarity on the equations' applicability to the problem at hand.
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Homework Statement
Why is the derivative assumed to be continuous?
Relevant Equations
dV_2/dn-dV_1/dn=sigma/epsilon
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1758149857826.webp


I don't understand how why the derivative is continuous. Clearly there is a surface charge and by gauss law we should have
1758149989980.webp
 
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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

Screen Shot 2025-09-17 at 11.03.11 PM.webp

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.
 
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