First slide

Electric dipole

Question

A neutral water molecule $(H_{2} 0)$ in its vapour state has an electric dipole moment of magnitude $6.4 \times 10^{- 30}$ C-m. How far apart are the molecules' centres of positive and
negative charge?

Moderate

A

4 m
B

4 mm
C

4 μm
D

4 pm

Solution

There are 10 electrons and 10 protons in a neutral water molecule.
So, its dipole moment is
p = q (21) = 10 e (21)
Hence, length of the dipole, i.e., distance between centres of positive and negative charges is
$2 l = \frac{p}{10 e} = \frac{6.4 \times 10^{- 30}}{10 \times 1.6 \times 10^{- 19}} = 4 \times 10^{- 12} m = 4 pm$

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