Question 6.13: A cylindrical capacitor has radii a=1 cm and b=2.5 cm. If th...

The procedure is the same as that taken in Section 6.5 except that eq. (6.27a)

V=-\int_{2}^{1}E\cdot dl=-\int_{b}^{a}\left[\frac{Q}{2\pi\varepsilon\rho L}a_{\rho}\right]\cdot d\rho a_{\rho}

now becomes

V=-\int_{b}^{a}\frac{Q}{2\pi\varepsilon_{o}\varepsilon_{r}\rho L}d\rho=-\frac{Q}{2\pi\varepsilon_{o} L}\int_{b}^{a}\frac{d\rho}{\rho\left(\frac{10+\rho}{\rho}\right)}=\frac{-Q}{2\pi\varepsilon_{o}L}\int_{b}^{a}\frac{d\rho}{10+\rho}

=\frac{-Q}{2\pi\varepsilon_{o}L}\ln(10+\rho)|_{b}^{a}=\frac{Q}{2\pi\varepsilon_{o}L}\ln\frac{10+b}{10+a}

Thus the capacitance per meter is (L=1m)

C=\frac{Q}{V}=\frac{2\pi\varepsilon_{o}}{\ln\frac{10+b}{10+a}}=2\pi\cdot\frac{10^{-9}}{36\pi}\cdot\frac{1}{\ln\frac{12.5}{11.0}}

C=434.6 pF/m