An air-spaced coaxial cable has an inner conductor $0.5 \mathrm{~cm}$ in diameter and an outer conductor $1.5 \mathrm{~cm}$ in diameter. When the inner conductor is at a potential of $+8000 \mathrm{~V}$ with respect to the grounded outer conductor, what is the charge per meter on the inner conductor?
Let the linear charge density for the inner conductor be $\lambda$. By symmetry we see that the field intensity at a point distance $r$ from the axis in the cable between the conductors is radial and its magnitude is given by Gauss' theorem as

$$
E=\frac{\lambda}{2 \pi \varepsilon_{0} r} .
$$

Then the potential difference between the inner and outer conductors is

$$
V=\int_{a}^{b} E d r=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln (b / a)
$$

with $a=1.5 \mathrm{~cm}, b=0.5 \mathrm{~cm}$, which gives

$$
\begin{aligned}
\lambda &=\frac{2 \pi \varepsilon_{0} V}{\ln (b / a)}=\frac{2 \pi \times 8.9 \times 10^{-12} \times 8000}{\ln (1.5 / 0.5)} \\
&=4.05 \times 10^{-7} \mathrm{C} / \mathrm{m} .
\end{aligned}
$$