The dipole moment of the water molecule (H2O) is 6.17×10−30C⋅m. Consider a water molecule located at the origin whose dipole moment p⃗ points in the positive x direction. A chlorine ion (Cl−), of charge −1.60×10−19C, is located at x=3.00×10−9 meters. Assume that this x value is much larger than the separation d between the charges in the dipole, so that the approximate expression for the electric field along the dipole axis can be used.
Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.

If the x value is much larger than the separation d between the charges in the dipole , the approximate expression for the electric field along the dipole axis

can be written as follows

E = 2kp / d³

k is 9 x 10⁹ , p is dipole moment and d is distance of the point concerned and middle of dipole.

Given p = 6.17 x 10⁻³⁰ Cm ,

d = 3 x 10⁻⁹ m

E = [tex]\frac{9\times10^9\times2\times6.17\times10^{-30}}{(3\times10^{-9})^3}[/tex]

= 4.113 X 10⁶ N/C

Force on charge 1.6 x 10⁻¹⁹ C

= E X 1.6 x 10⁻¹⁹

= 4.113 X 10⁶ X 1.6 x 10⁻¹⁹

= 6.58 X 10⁻¹³ N

lhmarianateixeira lhmarianateixeira · Answer 2 · 2021-11-29T19:54:46+01:00

Initially we must calculate the electrical force of the water molecule which is:

[tex]E= 6.58X10^{-13} N[/tex]

So, calculating the value of the electric field, it will be necessary to use the formula:

[tex]E=\frac{2KP}{D^{3} }[/tex]

Where K is a constant given by:

[tex]K=9X10^{9}[/tex]

P is the moment of the dipole and D is the distance from the point in the middle of two dipoles, so:

[tex]P= 6.17X10^{-30} C/m\\D= 3X10^{-9} m[/tex]

Performing Electric Field Calculation:

[tex]E= \frac{(9X10^{9})(2)(6.17X10^{-30})}{(3X10^{-9})^{3} }\\E= 4.113X10^{6} N/C\\[/tex]

Multiplying by the charge and in this way we will find only the value of the electric power:

[tex]= (E)(1.6X10^{-19})\\= 6.58X10^{-13} N[/tex]

Learn more: brainly.com/question/16260427