= Solution
Take the inner conductor to be at potential $V$ above the outer. A cylindrical <Gauss law> surface of length $\ell$ encloses charge $\lambda\ell$ and gives $E_r2\pi r\ell=\lambda\ell/\epsilon_0$. The potential difference is
$$
V=\int_a^bE_r\,dr=\frac{\lambda}{2\pi\epsilon_0}\ln\frac ba.
$$
Thus the <coaxial cylindrical capacitor> field and charge are
$$
\boxed{\lambda=\frac{2\pi\epsilon_0V}{\ln(b/a)},\qquad
\mathbf E=\frac{V}{r\ln(b/a)}\mathbf e_r.}
$$
The perfectly conducting surfaces have no tangential <electric field> in this idealized steady model.
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