Irodov 3.109: A Wire Facing Its Own Image – Amit & Deepali Sinha Foundation

Problem Irodov 3.109

A long straight wire of cross-sectional radius $a$ runs parallel to an infinite conducting plane, its axis a distance $b$ from the plane, with $a \ll b$ . Find the mutual capacitance between the wire and the plane per unit length of the wire.

The wire end on, radius a, its axis a distance b above the conducting plane (drawn schematically: the problem has a b). — The wire end on, radius $a$ , its axis a distance $b$ above the conducting plane (drawn schematically: the problem has $a \ll b$ ).

Given

wire radius $a$
axis-to-plane distance $b$
$a \ll b$

Find

the mutual capacitance per unit length

C

Solution

Step 1Replace the plane by an image

Put the charge $\lambda$ per unit length on the wire and $-\lambda$ on the plane. Since $a \ll b$ , the charge on the wire spreads almost uniformly round it, so outside the wire its field is that of a line charge $\lambda$ on its axis. The field above the plane is then the field of $\lambda$ together with its mirror image $-\lambda$ a distance $b$ below the plane: that pair makes the plane an equipotential, which is the condition the conductor imposes, and it has the right charge in the region above.

The axis charge λ of the wire and its image -λ at the mirror point a distance b below the surface; the lines that reach the plane meet it at right angles, on the induced negative charge; those leaving upward return to it outside the frame. — The axis charge $\lambda$ of the wire and its image $-\lambda$ at the mirror point a distance $b$ below the surface; the lines that reach the plane meet it at right angles, on the induced negative charge; those leaving upward return to it outside the frame.

Step 2The potential of the pair

For one line charge $\lambda$ , by symmetry $E$ is radial and the same all round, and Gauss’s theorem on a coaxial cylinder of radius $r$ and length $\ell$ gives

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

The potential follows from $E = -d\varphi/dr$ , integrating the field over $r$ :

\varphi(r) = -\int\frac{\lambda\,dr}{2\pi\varepsilon_0 r} = -\frac{\lambda}{2\pi\varepsilon_0}\ln r + \text{const}

At a point at distance $r_1$ from the wire’s axis and $r_2$ from the image ( $-\lambda$ , so its term has the opposite sign), the two add:

\varphi = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{r_2}{r_1} \tag{1}

The additive constant has been set so that $\varphi = 0$ on the plane, where every point has $r_1 = r_2$ .

Step 3The voltage between wire and plane

On the wire’s surface $r_1 = a$ , while the distance to the image runs from $2b - a$ to $2b + a$ round the wire, which is $2b$ to within the small fraction $a/2b$ . Substituting in (1):

V = \varphi_{\mathrm{wire}} - 0 = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{2b}{a}

Step 4The 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{2\pi\varepsilon_0}{\ln(2b/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, as asked. Against the exact result: for a wire of any thickness the wire’s surface is still an equipotential of $\lambda$ and its image, provided the line charge sits not on the axis but at the inverse point, a distance $c$ from the plane with $c^2 = b^2 - a^2$ ; the potential (1) on that surface then gives the capacitance $2\pi\varepsilon_0/\operatorname{arccosh}(b/a)$ . For $b \gg a$ ,

\operatorname{arccosh}\frac{b}{a} = \ln\left(\frac{b}{a} + \sqrt{\frac{b^2}{a^2} - 1}\right) \approx \ln\frac{2b}{a},

which turns the exact form into (2).

Remark

The wire over a plane holds exactly twice the capacitance per unit length of the two-wire line with axes $2b$ apart: the plane sits on the midplane of that line, where its potential is zero, so the wire to plane voltage is half the line’s voltage for the same $\lambda$ . The same image argument gives the capacitance of an overhead power line to the ground.