Question

A water molecule consists of two hydrogen atoms bonded with one oxygen atom. The bond angle between the two hydrogen atoms is 104° (see below). Calculate the net dipole moment (in C · m) of a water molecule that is placed in a uniform, horizontal electric field of magnitude 5.7 ✕ 10−8 N/C. (Only calculate the permanent dipole moment based on charge distributions shown in the figure, and exclude any induced dipole moment. The O–H bond length in water is 0.958 angstroms. Express your answer in vector form. Assume that the +x-axis is to the right and the +y-axis is up along the page.)

Answer 1

Explanation:

According to the figure,

                    $\vec{p} = q\vec{d}$

Formula to calculate net dipole moment is as follows.

          $\vec{p_{net}} = 2P Cos 52^{o}\hat{i} + (0)\hat{j}$

           $\vec{p_{net}} = 2qd Cos 52^{o}\hat{i}$

                        = $2 \times 1.6 \times 10^{-19} \times 0.958 \times 10^{-10} \times Cos 52^{o}\hat{i}$

                        = $1.88 \times 10^{-29}\hat{i}$ C-m

Therefore, we can conclude that the net dipole moment for given water molecule is $1.88 \times 10^{-29}\hat{i}$ C-m.