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[SOLVED] Electric potential Laplace


Well-known member
Feb 1, 2012
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 \(n\).
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}A(k)\frac{\delta(k - k')}{k}dk && = {}
A(k) &= kV_0\int_0^{\infty}r\mathcal{J}_0(k'r)dk
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?
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