What does it mean when two point charges form a dipole a small distance away from each other?

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What does it mean when two point charges form a dipole a small distance away from each other?

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Answer 1 · 2015-07-06T16:18:28

Small, meaning close enough together.

Two charges can only interact well enough to form a dipole if they are close enough together. Too far apart and they are essentially independent charges (does not mean they aren't interacting).

Consider the coulombic interaction between two point charges:

$V (\to r) = \frac{q_{1} q_{2}}{4 π ε_{0} {\to r}_{12}}$ $V(vecr) = (q_1 q_2)/(4piepsilon_0vecr_(12)$

where $q_{i}$ $q_i$ is the charge of particle $i$ $i$ , $ε_{0} = 8.854 \times 10^{- 12} F \cdot m^{- 1}$ $epsilon_0 = 8.854 xx 10^(-12) "F"cdot"m"^(-1)$ is the vacuum permittivity constant, and ${\to r}_{12}$ $vecr_(12)$ is the radial separation between two particles.

This means the coulombic potential of the two charges depends on the distance between them and the charges' magnitudes. Large $r$ $r$ , small potential. Large enough potential, and a dipole forms, where a dipole between two point charges of charge $\pm q$ $pmq$ is defined as:

${\to μ}_{12} = q \cdot {\to r}_{12}$ $vecmu_(12) = qcdotvecr_(12)$

An example of a small charge is $e^{-} = 1.6022 \times 10^{- 19} C$ $e^(-) = 1.6022 xx 10^(-19) "C"$ .

An example of "small" for molecules is a few $o A$ $stackrel("o")("A")$ . The bond length of the $OH$ $"OH"$ bond in $H_{2} O$ $"H"_2"O"$ is $0.958$ $0.958$ $o A$ $stackrel("o")("A")$ , where $1$ $1$ $o A = 10^{- 10} m$ $stackrel("o")("A") = 10^(-10) "m"$ is an angstrom.

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What does it mean when two point charges form a dipole a small distance away from each other?

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