Question:
A neutral water molecule $(H_2O)$ 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 charges?
A neutral water molecule $(H_2O)$ 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 charges?
Updated On: Apr 15, 2024
- 4m
- 4 mm
- $4 \mu m$
- 4:00 PM
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The Correct Option is D
Solution and Explanation
There are 10 electrons and 10 protons in a neutral water molecule.
So, its dipole moment is $p=q(2 l)=10 e(2 l)$
Hence length of the dipole ie, distance between centres of positive and negative charges is
$2l=\frac{p}{10 e} =\frac{6.4 \times 10^{-30}}{10 \times 1.6 \times 10^{-19}}$
$=4 \times 10^{-12} m$
$=4\, pm$
So, its dipole moment is $p=q(2 l)=10 e(2 l)$
Hence length of the dipole ie, distance between centres of positive and negative charges is
$2l=\frac{p}{10 e} =\frac{6.4 \times 10^{-30}}{10 \times 1.6 \times 10^{-19}}$
$=4 \times 10^{-12} m$
$=4\, pm$
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Concepts Used:
Electric Dipole
An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance of 2a. The direction from q to -q is said to be the direction in space.
p=q×2a
where,
p denotes the electric dipole moment, pointing from the negative charge to the positive charge.
Force Applied on Electric Dipole
