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Please prove $$1\ debye = 10^{-18} \ esu \ cm = 3.335 \times 10^{30} \ C - m$$ ( coulomb meter),

Solution
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Dipole moment $$(\mu)$$ between two charges separated by distance $$d$$ is
$$\mu = q \times d$$

Historically the debye was defined as the dipole moment resulting from two charges of opposite sign but an equal magnitude of $$10^{-10} $$ statcoulomb (generally called e.s.u. (electrostatic unit) in older literature), which were separated by $$1$$ angstrom $$(= 10^{-8}\ cm)$$

$$1$$ Debye = $$10^{-10} esu \times 10^{-8}\ cm $$

$$ = 10^{-18}\ esu\ cm $$

$$ = 10^{-18}\ esu\ cm \times \dfrac{3.335 \times 10^{-10}\ C}{\text{esu}} \times \dfrac{10^{-2}\ m}{cm} $$

$$ = 3.335 \times 10^{-30}\ cm $$

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