How Does Electric Potential Behave Across a Circular Boundary in Electrostatics?

[Dustinsfl]

The inside of a circle of radius \(R\) is held at a constant potential \(V_0\) in the xy plane. The outside is held at a constant potential \(0\).

Show that that the electric static potential for \(z > 0\) is
\[
T(r, z) = V_0R\int_0^{\infty}\mathcal{J}_0(kr)\mathcal{J}_1(kR)e^{-kz}dk.
\]
Since we need boundedness as \(z\to\infty\), \(Z(z)\sim e^{-kz}\); additionally, since we need boundedness at the origin, \(R(r)\sim\mathcal{J}_0(kr)\) where \(\mathcal{J}_n\) is the bessel function of order \(n\).
\begin{alignat}{2}
T(r, z) &= \int_0^{\infty}A(k)\mathcal{J}_0(kr)e^{-kz}dk\\
T(r, 0) &= \int_0^{\infty}A(k)\mathcal{J}_0(kr)dk && ={} V_0\\
& \int_0^{\infty}\int_0^{\infty}A(k)r\mathcal{J}_0(kr)\mathcal{J}_0(k'r)dk && = {}
\int_0^{\infty}V_0\mathcal{J}_0(k'r)dk\\
& \int_0^{\infty}A(k)\frac{\delta(k - k')}{k}dk && = {}
\int_0^{\infty}rV_0\mathcal{J}_0(k'r)dk\\
A(k) &= kV_0\int_0^{\infty}r\mathcal{J}_0(k'r)dk
\end{alignat}
However, I need to get that \(A(k) = V_0R\mathcal{J}_1(kR)\).

Now if I do the same integration but the RHS from \(r\in[0, R]\), I get
\[
A(k) = V_0R\mathcal{J}_1(kR)
\]
which is correct. How do I justify the integration on the LHS to infinity but the RHS only to \(R\)? Simply because outside of \(R\) the potential is \(0\) but why wouldn't that be the bounds for the LHS then?

 

[jvicens]

I can help you justify the integration on the LHS to infinity. It is important to note that the potential is held constant at \(V_0\) for all values of \(r\) within the circle of radius \(R\). This means that for values of \(r\) outside of the circle, the potential is also constant at \(V_0\). Therefore, the integral on the LHS can be extended to infinity because the potential outside of the circle is also constant at \(V_0\). This is not the case for the RHS, as the potential outside of the circle is \(0\), not \(V_0\). This is why the integration on the RHS is only extended to \(R\).