= Solution
Use opposite <electric charges> $\pm Q$ over a common <length> $L$, and write the inner <linear charge density> as $\lambda=Q/L$. Away from the ends, <axisymmetry> and <Gauss's law> on a coaxial cylindrical <Gaussian surface> of <length> $\ell$ give
$$
E(r)\,2\pi r\ell=\frac{\lambda\ell}{\epsilon_0},\qquad
E(r)=\frac{\lambda}{2\pi\epsilon_0r},\qquad a<r<b.
$$
The <electric potential difference> and <capacitance> of the <coaxial cylindrical capacitor> are
$$
V=\int_a^bE(r)\,dr=\frac Q{2\pi\epsilon_0L}\log\frac ba,
\qquad\boxed{C=\frac{2\pi\epsilon_0L}{\log(b/a)}.}
$$
This neglects the <fringing field of a capacitor> at the ends, as justified when $L\gg b$.
Back to article page
