A neutral water molecule H2O in its vapor state has an electric dipole moment of magnitude 6-4 -10-30 C-m- How far apart are the centres of positive and negative charge-

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Dipole Moment

A neutral wat...

Question

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

4 m

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4 m m

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4 μ m

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4 p m

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Solution

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

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A neutral water molecule (H2O) in its vapor state has an electric dipole moment of magnitude 6.4×10–30 C-m. How far apart are the centres of positive and negative charge?

The correct option is D 4pmThere are 10 electrons and 10 protons in a neutral water molecule. So it's dipole moment is p = q (2l) = 10 e (2l) Hence length of the dipole i.e. distance between centres of positive and negative charges is 2l=p10e=6.4×10−3010×1.6×10−19=4×10−12m=4 pm

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