# [SOLVED]Electric potential Laplace

#### dwsmith

##### Well-known member
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?

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