Please prove $1 d e b y e = 1 0^{- 18} e s u c m = 3.335 \times 1 0^{30} C - m$ ( coulomb meter),

Question

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Answer 1

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

Historically the debye was defined as the dipole moment resulting from two charges of opposite sign but an equal magnitude of $1 0^{- 10}$ statcoulomb (generally called e.s.u. (electrostatic unit) in older literature), which were separated by $1$ angstrom $(= 1 0^{- 8} c m)$

$1$ Debye = $1 0^{- 10} e s u \times 1 0^{- 8} c m$

$= 1 0^{- 18} e s u c m$

$= 1 0^{- 18} e s u c m \times \frac{3 . 3 3 5 \times 1 0 ^{- 10} C}{esu} \times \frac{1 0 ^{- 2} m}{c m}$

$= 3.335 \times 1 0^{- 30} c m$

Answer 2

Dipole moment

(μ)

between two charges separated by distance

d

is

μ = q \times d

Historically the debye was defined as the dipole moment resulting from two charges of opposite sign but an equal magnitude of

1 0^{- 10}

statcoulomb (generally called e.s.u. (electrostatic unit) in older literature), which were separated by

1

angstrom

(= 1 0^{- 8} c m)

1

Debye =

1 0^{- 10} e s u \times 1 0^{- 8} c m

= 1 0^{- 18} e s u c m

= 1 0^{- 18} e s u c m \times \frac{3 . 3 3 5 \times 1 0 ^{- 10} C}{esu} \times \frac{1 0 ^{- 2} m}{c m}

= 3.335 \times 1 0^{- 30} c m