If C1 & C2 and P1 & P2 are the pairs of current and potential electrodes in the Wenner (W) and Dipole-Dipole (DD) array configurations as shown in the figure below, then the fraction of the geometric factor for DD array that will be equal to half of that of W array is _______________ (rounded off to three decimal places). (Use n = 1 in DD array)
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Correct Answer: 0.166
Solution and Explanation
Step 1 - Geometric factor of the Wenner array: With equal electrode spacing \(a\) between C1-P1, P1-P2 and P2-C2, the Wenner geometric factor is
\[K_W = 2\pi a\]
Step 2 - Geometric factor of the Dipole-Dipole array: With current-dipole spacing \(a\), potential-dipole spacing \(a\), and dipole separation factor \(n\) (the gap between the two dipoles is \(na\)), the standard Dipole-Dipole geometric factor is
\[K_{DD} = \pi n(n+1)(n+2)a\]
For \(n=1\):
\[K_{DD} = \pi(1)(2)(3)a = 6\pi a\]
Step 3 - Set up the required fraction: We need the fraction \(x\) of \(K_{DD}\) that equals half of \(K_W\):
\[x\cdot K_{DD} = \frac{1}{2}K_W\]
\[x = \frac{K_W/2}{K_{DD}} = \frac{2\pi a/2}{6\pi a} = \frac{\pi a}{6\pi a} = \frac{1}{6}\]
Step 4: Evaluating,
\[x = \frac{1}{6} = 0.1\overline{6} \approx 0.167\]
This lies within the accepted range (0.166 to 0.167).
\(\boxed{x\approx 0.167}\)
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