Irodov 3.108: Two Wires, One Logarithm – Amit & Deepali Sinha Foundation

Problem Irodov 3.108

Two long straight parallel wires in air have the same cross-sectional radius $a$ , and their axes are a distance $b$ apart, with $b \gg a$ . Find the mutual capacitance of the pair per unit length of the wires.

The two wires end on, each of radius a, with their axes a distance b apart (drawn schematically: the problem has b a). — The two wires end on, each of radius $a$ , with their axes a distance $b$ apart (drawn schematically: the problem has $b \gg a$ ).

Given

wire radius $a$
axis separation $b \gg a$
air

Find

the mutual capacitance per unit length

C

Solution

Step 1The field between the wires

Put the charges $\lambda$ and $-\lambda$ per unit length on the two wires. Since $b \gg a$ , each wire barely disturbs the charge on the other, so the charge on each stays almost uniform round it and acts outside it like a line charge on its axis. For one line charge, by symmetry $E$ is radial and the same all round, so

E\cdot 2\pi r\ell = \frac{\lambda\ell}{\varepsilon_0}

The wires charged λ and -λ, and a point P on the line of centers at distance x from the positive wire's axis; both wires make the field there point toward the negative wire. — The wires charged $\lambda$ and $-\lambda$ , and a point $P$ on the line of centers at distance $x$ from the positive wire’s axis; both wires make the field there point toward the negative wire.

At a point $P$ on the line of centers, a distance $x$ from the positive wire’s axis and $b - x$ from the negative one’s, both contributions point toward the negative wire and add:

E(x) = \frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{x} + \frac{1}{b - x}\right) \tag{1}

Step 2The voltage between the wires

The voltage is the integral of (1) along the line of centers, from the surface of one wire, $x = a$ , to the surface of the other, $x = b - a$ :

\begin{aligned} V &= \frac{\lambda}{2\pi\varepsilon_0}\int_a^{b-a}\left(\frac{1}{x} + \frac{1}{b - x}\right)dx = \frac{\lambda}{2\pi\varepsilon_0}\left[\ln\frac{x}{b - x}\right]_a^{b-a}\\ &= \frac{\lambda}{2\pi\varepsilon_0}\cdot 2\ln\frac{b - a}{a} = \frac{\lambda}{\pi\varepsilon_0}\ln\frac{b - a}{a} \end{aligned}

For $b \gg a$ , $b - a$ may be replaced by $b$ :

V \approx \frac{\lambda}{\pi\varepsilon_0}\ln\frac{b}{a}

Step 3The capacitance per unit length

The mutual capacitance per unit length is the charge per unit length over the voltage, $C = \lambda/V$ :

C \approx \dfrac{\pi\varepsilon_0}{\ln(b/a)}

(2)

Check

Units: $\varepsilon_0$ is in F/m and the logarithm is a pure number, so $C$ is a capacitance per unit length. Trend: moving the wires apart (larger $b$ ) or thinning them (smaller $a$ ) must lower the capacitance, since the same charge then needs a larger voltage, and the logarithm in the denominator grows in both cases. Against the exact treatment, in which the two cylinders are equipotentials of a pair of line charges $\pm\lambda$ placed not on the axes but at the inverse points, a distance $c$ from the midpoint with $c^2 = (b/2)^2 - a^2$ , and which gives $\pi\varepsilon_0/\operatorname{arccosh}(b/2a)$ : for $b \gg a$ , $\operatorname{arccosh}(b/2a) \approx \ln(b/a)$ , the same result.

Remark

The capacitance depends on the geometry only through the logarithm of $b/a$ , so it is remarkably insensitive to it: moving the wires ten times farther apart changes $C$ only by the ratio of $\ln(10b/a)$ to $\ln(b/a)$ . For a typical two-wire line, $b/a$ of order 100, $C$ is about 6 pF per meter, or 6 nF per kilometer of line.