Four metallic measuring tips are contacted to a thin foil of thickness \(d\) and conductivity \(\sigma\). A current source is connected to the two outer tips \((I_+\) and \(I_-)\) which transmit a constant current \(I\) through the sample. A voltage is measured at the two inner tips \((U_+\) and \(U_-).\)

Which formula describes the electrical resistance \[R = \frac{U_+ - U_-}{I} \] in this geometry?

**Details and Assumptions:**

- The foil has an almost infinite size \((L \gg l)\) but a very small thickness \((d \ll l).\)
- The tips are equidistant with a distance of \(l\) between the outer tips. The contact point is infinitely small and the voltage measurement does not affect the electric fields inside the foil.
- The points \(I_+ \text{ and } I_-\) can be treated as the point source and point drain, respectively, for the planar current density \(\vec j(x,y).\)

**Important Relations:**

- Current Density (Ohm's law): \[ \vec j (\vec r) = \sigma \cdot \vec E (\vec r) \ \ \text{with} \ \ I = \int_A \vec j (\vec r) \cdot d\vec A, \] where \(\vec E\) is the electric field and \(I\) is the current flow through the plane \(A\).
- Electric Voltage: \[ U_{12} = \phi(\vec r_2) - \phi(\vec r_1) = -\int_1^2 \vec E (\vec r) \cdot d\vec r, \] measured between points \(\vec r_1\) and \(\vec r_2\) with the electric potential \(\phi(\vec r)\).

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