Irodov 3.31: Twice as Far, a Logarithm – Amit & Deepali Sinha Foundation

Problem Irodov 3.31

An infinitely long straight thread carries charge $\lambda = 0.40$ $\mu$ C/m per unit length. Point $2$ is $\eta = 2.0$ times as far from the thread as point $1$ . Find the potential difference between points $1$ and $2$ .

The charged thread and the two points: point 1 at distance r_1 from it, point 2 at η r_1 (drawn for η = 2). — The charged thread and the two points: point 1 at distance $r_1$ from it, point 2 at $\eta r_1$ (drawn for $\eta = 2$ ).

Given

$\lambda = 0.40$ $\mu$ C/m
$\eta = 2.0$

Find

\varphi_1 - \varphi_2

Solution

Step 1The field of the thread

By symmetry the field is radial and the same all round a coaxial cylinder of radius $r$ and length $\ell$ , so Gauss’s law gives

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

which is

E = \frac{\lambda}{2\pi\varepsilon_0 r} \tag{1}

Step 2The potential difference as a line integral

Let point 1 be at distance $r_1$ from the thread and point 2 at $r_2 = \eta r_1$ . The potential difference is the integral of the field along any path from 1 to 2; along a radial path, with (1):

\begin{aligned} \varphi_1 - \varphi_2 &= \int_{r_1}^{\eta r_1} E\,dr = \frac{\lambda}{2\pi\varepsilon_0}\int_{r_1}^{\eta r_1}\frac{dr}{r}\\ &= \frac{\lambda}{2\pi\varepsilon_0}\,\ln\frac{\eta r_1}{r_1} \end{aligned}

The distance $r_1$ cancels, and with $1/(4\pi\varepsilon_0) = 9.0\times10^{9}$ m/F the number is

\begin{aligned} \varphi_1 - \varphi_2 &= 2\,(9.0\times10^{9})(0.40\times10^{-6})\ln 2.0\\ &= (7.2\times10^{3})(0.693) \text{ V} \end{aligned}

\varphi_1 - \varphi_2 = \frac{\lambda}{2\pi\varepsilon_0}\ln\eta \approx 5.0 \text{ kV}

(2)

Point 1, nearer the positively charged thread, is at the higher potential.

Check

Units: $\lambda/\varepsilon_0$ is (C/m)/(F/m) $=$ C/F $=$ V, and the logarithm is a pure number. With $\eta = 1$ the two points are equally far and the difference is zero, as it must be; with $\eta < 1$ the logarithm turns negative, and point 2, now the nearer one, is the higher, as the sign of the field requires.

Remark

Only the ratio of the distances enters, so the potential difference between radii $1$ mm and $2$ mm is the same as between $1$ km and $2$ km. This is the logarithmic potential of two-dimensional electrostatics, and it is why the capacitance per unit length of a coaxial cable depends only on the ratio of its two radii.