Question

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)

Hint:

Write K_W = 2*pi*a and K_DD = pi*n(n+1)(n+2)*a; with n = 1, K_DD = 6*pi*a, so the required fraction is (K_W/2)/K_DD = 1/6.

Answer 1

Correct Answer: 0.166

An equivalent, quicker route is to first form the ratio of the two geometric factors directly, then rearrange as needed.

\[\frac{K_{DD}}{K_W} = \frac{\pi n(n+1)(n+2)a}{2\pi a} = \frac{n(n+1)(n+2)}{2}\]

For \(n=1\): \(\dfrac{K_{DD}}{K_W} = \dfrac{1\cdot2\cdot3}{2} = 3\), i.e. \(K_{DD} = 3K_W\).

The question asks for the fraction \(x\) of \(K_{DD}\) equal to HALF of \(K_W\), i.e. \(xK_{DD} = 0.5K_W\). Since \(K_{DD}=3K_W\),

\[x(3K_W) = 0.5K_W \implies x = \frac{0.5}{3} = \frac{1}{6}\approx 0.167\]

Both routes agree: the geometric factor of the Dipole-Dipole array (with n = 1) is 6 times as large as the Wenner array's for the same unit electrode spacing \(a\), so only 1/6 of \(K_{DD}\) is needed to match half of \(K_W\).

\(\boxed{x\approx 0.167}\), matching the accepted range 0.166 to 0.167.