- #1

yucheng

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- Homework Statement
- Two coaxial, nonconducting cylinders have surface charge densities ##\sigma_a(\phi)## on ##\sigma_b(\phi)## on the inner and outer cylinders, giving rise to potentials ##V_a(\phi)## and ##V_b(\phi)## on the two surfaces. Find the potential (a) for ##r<a## (b) ##r>a## and (c) for ##r \in (a,b)##

(Vanderlinde, Example 5.3)

- Relevant Equations
- N/A

Hi!

The problem clearly states that there is a surface charge density, which somehow gives rise to a potential.

The author has solved the Laplace equation in cylindrical coordinates and applied the equation to the problem.

So ##\nabla^2 V(r,\phi) = 0##, and ##V(a,\phi) = V_a(\phi)## (where the potential takes on this value at the boundary). However, because of the surface charge at ##r=a##, doesn't the potential also satisfy Poisson's equation? Does this mean that given, ##V_a(\phi)##, we can also find ##\sigma_a##?

I cooked up an explanation to this:

$$\nabla^2 V(r,\phi) = \frac{\rho_a}{\epsilon_0} = \frac{ \sigma_a \delta(r-a)}{\epsilon_0}$$

But this is infinite at ##r=a## what's wrong? Is it because ##V(r,\phi)## is not valid at the boundary? I know it solves Laplace equation hence not valid where there are charges (that is why in the problem statement, the author only solves for ##r<a##). What does ##V(a,\phi) = V_a(\phi)## mean then?

Thanks in advance!

P.S. looks like someone asked this before but never got a reply...

https://www.physicsforums.com/threads/electromagnetics-parallel-plates-poisson-laplace.134235/

The problem clearly states that there is a surface charge density, which somehow gives rise to a potential.

The author has solved the Laplace equation in cylindrical coordinates and applied the equation to the problem.

So ##\nabla^2 V(r,\phi) = 0##, and ##V(a,\phi) = V_a(\phi)## (where the potential takes on this value at the boundary). However, because of the surface charge at ##r=a##, doesn't the potential also satisfy Poisson's equation? Does this mean that given, ##V_a(\phi)##, we can also find ##\sigma_a##?

I cooked up an explanation to this:

$$\nabla^2 V(r,\phi) = \frac{\rho_a}{\epsilon_0} = \frac{ \sigma_a \delta(r-a)}{\epsilon_0}$$

But this is infinite at ##r=a## what's wrong? Is it because ##V(r,\phi)## is not valid at the boundary? I know it solves Laplace equation hence not valid where there are charges (that is why in the problem statement, the author only solves for ##r<a##). What does ##V(a,\phi) = V_a(\phi)## mean then?

Thanks in advance!

P.S. looks like someone asked this before but never got a reply...

https://www.physicsforums.com/threads/electromagnetics-parallel-plates-poisson-laplace.134235/

Last edited: