Four Point Probe

A single radial current point source at the origin can be described by $\vec j(\vec r) = j(r) \vec e_r \quad \text{with} \quad \vec e_r = \left( \begin{array}{c} \cos \varphi \\ \sin \varphi \end{array} \right) = \frac{1}{r} \vec r$ with the help of polar coordinates $r$ (radial distance) and $\varphi$ (polar angle). Therefore, the absolute value $j(r)$ of the current density depends only on the radial distance from its source and the unit vector $\vec e_r$ points radial outwards. The vector field $\vec j(\vec r)$ has this form because of its radial symmetry. The surface integral $\int_{r = \text{const}} \vec j(\vec r) \cdot d\vec A = j(r) \cdot 2 \pi r d \stackrel{!}{=} I$ over a cylinder with radius $r$ and height $d$ around the origin yields the total electric current $I$ . Therefore, $j(r) = \frac{I}{2 \pi d r}$ We assume two points sources at the points $- \frac{1}{2} l \vec e_x$ and $\frac{1}{2} l \vec e_x$ with a radial distance $r_1 =|\vec r + \frac{1}{2} l \vec e_x| \quad \text{ and } \quad r_2 = |\vec r - \frac{1}{2} l \vec e_x|$ and a radial direction $\vec e_{r1} =\frac{\vec r + \frac{1}{2} l \vec e_x}{|\vec r + \frac{1}{2} l \vec e_x|} \quad \text{ and } \quad \vec e_{r3} =\frac{\vec r - \frac{1}{2} l \vec e_x}{|\vec r - \frac{1}{2} l \vec e_x|}$ The total current density yields $\vec j(x, y) = \frac{I}{2 \pi d} \left[ \frac{\vec r + \frac{1}{2} l \vec e_x}{|\vec r + \frac{1}{2} l \vec e_x|^2} - \frac{\vec r - \frac{1}{2} l \vec e_x}{|\vec r - \frac{1}{2} l \vec e_x|^2} \right]$ Calculating the path integral from $-\frac{1}{6} l \vec e_x$ to $+\frac{1}{6} l \vec e_x$ along the x-direction, we get $U = \int_{-\frac{1}{6} l \vec e_x}^{\frac{1}{6} l \vec e_x} \frac{1}{\sigma} \vec j(\vec r) \cdot d\vec r = \frac{I}{2 \pi \sigma d} \int_{-l/6}^{l/6} \left[ \frac{1}{x + l/2} - \frac{1}{x - l/2} \right] dx = \frac{\ln 2}{\pi \sigma d}$