Capacitance of a single straight wire What is the capacitance of a single straight wire?
calculating the electric field using Gauss's law, I get a constant divided by the distance from the wire (r). Integrating 1/r gives me ln(r). evaluating r at the radius of the wire and infinity gives me infinity. On the one hand, I am thinking that I am using the field from an infinitely long straight wire, maybe an infinitely long wire has infinite capacitance. But what is confusing is that using q = CV (charge = capacitance times voltage) , I get cL = Cck*ln(r) where c is charge per unit length and k is the rest of the constants. There I am plugging in L, which means I am not technically using an infinitely long wire.
 A: Using Gauss's law, you should have found that the field strength (radial) at distance $r$ from the central axis of a long straight wire of length $\ell$ and radius $r_1$ carrying charge $Q$ is of magnitude
$$E= \frac{Q}{4 \pi \epsilon_0 \ell r}$$
So the pd between the surface of the wire (of radius $r_1$) and a surrounding co-axial conducting surface of radius $r_2$ is
$$V=\int_{r_1} ^{r_2} \frac{Q}{4 \pi \epsilon_0 \ell r}dr=\frac{Q}{4 \pi \epsilon_0 \ell} \ln \frac{r_2}{r_1}$$
The capacitance is therefore
$$C=\frac QV=\frac{4 \pi \epsilon_0 \ell}{\ln \frac{r_2}{r_1}}$$
So, not surprisingly, the capacitance is proportional to the length of the wire.
More interestingly, the capacitance goes to zero as we make the surrounding conducting surface larger and larger ($r_2>>r_1$). In other words, an isolated conducting wire would have zero capacitance. In practice there will be objects at various distances from the wire, and the charged wire will induce charges on these objects, so the system's capacitance, though hard to calculate, will not actually be zero (when it can be defined at all).
Hope this helps.
