For all following sections, assume |2|= 1 |p|= 5, R₁ = 5 mm, R₁ =15 mm, and R₂ = 2 cm. Consider a water molecule (H₂O). The two hydrogen atoms are positively charged, the oxygenatom is doubly negatively charged. The bond angle between the hydrogen atoms is 104.5°. Theoxygen-hydrogen distance is 0.096 nm.104.5"0.096 nmT:Considering the atoms as non-bonded, independent charges... what is the netelectric force vector on the oxygen atom (blue)?insidering the atoms as non-bonded, independent charges... what is thepotential energy of this arrangement?Iculate the electric dipole moment (i.e. dipole vector) of the water molecule,using the axes in the figure to establish unit vectors. Hint: Treat the molecule as twoindependent dipoles, and add the moments as vectors.I claim it is p = -1.88 x 10-299 Cm. Am I correct? Why is it negative?(NOTE: The measured value of the dipole moment of water is about [p] =0.617 x 10-29 C m, or about 1/3 of our calculated value.

Given:charge on hydrogen atoms, qh = +e = +1.6×10−19 Ctotal charge on oxygen atom, qo = −2e =…

Answered: Consider a water molecule (H₂O). The…

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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For all following sections, assume |2|= 1 |p|= 5, R₁ = 5 mm, R₁ =
15 mm, and R₂ = 2 cm.

Consider a water molecule (H₂O). The two hydrogen atoms are positively charged, the oxygen
atom is doubly negatively charged. The bond angle between the hydrogen atoms is 104.5°. The
oxygen-hydrogen distance is 0.096 nm.
104.5"
0.096 nm
T:
Considering the atoms as non-bonded, independent charges... what is the net
electric force vector on the oxygen atom (blue)?
insidering the atoms as non-bonded, independent charges... what is the
potential energy of this arrangement?
Iculate the electric dipole moment (i.e. dipole vector) of the water molecule,
using the axes in the figure to establish unit vectors. Hint: Treat the molecule as two
independent dipoles, and add the moments as vectors.
I claim it is p = -1.88 x 10-299 Cm. Am I correct? Why is it negative?
(NOTE: The measured value of the dipole moment of water is about [p] =
0.617 x 10-29 C m, or about 1/3 of our calculated value.

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$\begin{array}{l} G i v e n : \\ c h a r g e o n h y d r o g e n a t o m s, q_{h} = + e = + 1.6 \times 10^{- 19} C \\ t o t a l c h a r g e o n o x y g e n a t o m, q_{o} = - 2 e = - 3.2 \times 10^{- 19} C \\ h y d r o g e n - o x y g e n d i s t a n c e, r = 0.096 n m = 0.096 \times 10^{- 9} m \\ a n g l e b e t w e e n t h e H - O b o n d, θ = {104.5}^{o} \\ A s k e d : \\ i . N e t e l e c t r i c f o r c e v e c t o r o n o x y g e n a t o m, \vec{F} = ? \\ i i . P o t e n t i a l e n e r g y o f t h e s y s t e m, U = ? \\ i i i . n e t e l e c t r i c d i p o l e o f t h e m o l e c u l e, \vec{p} = ? \end{array}$

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