Question

Water is a polar molecule. It has an electric dipole strength of p-6-10 30 C-n. This is roughly equivalent to a proton and an electron separated by a distance of d 4-10-11 m (about the radius of a hydrogen atom) In this problem, we will use Coulomb's law to compare the force between two protons, a proton and a water molecule, and two water molecules. Use the arrangement of the charges shown in the figure for your calculations. (a) What is the force between two protons separated by a distance of r-0.1 nm? (b) What is the force between a proton and a water molecule separated by a distance of r 0.1 nm? (c) What is the force between two water molecules separated by a distance of r0.1 nm? (d) Repeat the analysis of parts (a) (c) for a separation of r = 1 nm (e) Critical Thinking: You should now have six differ- ent numbers. What do they tell you? Compare ratios of various pairs of numbers and comment on the relative strength of the interactions (charge-charge, charge-dipole, dipole-dipole) and how these depend on distance.

Answer 1

a)

Force between two protons separated by distance $r=0.1\,n m=0.1*10^{-9}\,m$

$F_{p,p}=\frac{e*e}{4\pi\epsilon_0 r^2}=8.99*10^9*\frac{(1.6*10^{-19})^2}{(0.1*10^{-9})^2}=2.30*10^{-8}\,N$

b)

$r=0.1\,n m=0.1*10^{-9}\,m$

Electric field at the location of proton due to water molecule (dipole ) is $E_{p,dp}=\frac{2p}{4\pi\epsilon_0 r^3}$

Force between a proton and a water molecule is $F_{p,dp}=eE_{p,dp}=\frac{2pe}{4\pi\epsilon_0 r^3}$

$F_{p,dp}=8.99*10^9*\frac{2*6*10^{-30}*1.6*10^{-19}}{(0.1*10^{-9})^3}=1.73*10^{-8\,N}$

c)

$r=0.1\,n m=0.1*10^{-9}\,m$

Electric field at the location of a dipole due to another dipole is $E_{dp,dp}=\frac{2p}{4\pi\epsilon_0 r^3}$

Force on one dipole due to other is $|F_{dp,dp}|=p\frac{d}{dr}(E_{dp,dp})=p\frac{d}{dr}\left ( \frac{2p}{4\pi\epsilon_0 r^3} \right )=|-\frac{6p^2}{4\pi\epsilon_0 r^4}|=\frac{6p^2}{4\pi\epsilon_0 r^4}$

$F_{dp,dp}=\frac{6p^2}{4\pi\epsilon_0 r^4}$

$F_{dp,dp}=8.99*10^9*\frac{6(6*10^{-30})^2}{(0.1*10^{-9})^4}=1.94*10^{-8}\,N$

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Repeating above calculations for  $r=1\,n m=1*10^{-9}\,m$

a)

Force between two protons is

$F_{p,p}=\frac{e*e}{4\pi\epsilon_0 r^2}=8.99*10^9*\frac{(1.6*10^{-19})^2}{(1*10^{-9})^2}=2.30*10^{-10}\,N$

b)

Force between a proton and a water molecule is $F_{p,dp}=eE_{p,dp}=\frac{2pe}{4\pi\epsilon_0 r^3}$

$F_{p,dp}=8.99*10^9*\frac{2*6*10^{-30}*1.6*10^{-19}}{(1*10^{-9})^3}=1.73*10^{-11\,N}$

c)Force on one dipole due to other is

$F_{dp,dp}=\frac{6p^2}{4\pi\epsilon_0 r^4}$

$F_{dp,dp}=8.99*10^9*\frac{6(6*10^{-30})^2}{(1*10^{-9})^4}=1.94*10^{-12}\,N$

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e)

From the above calculations,

As the distance between the charge-charge , charge-dipole and dipole-dipole is increasing, strength of electric force between the charges and dipoles is decreasing. With distance electric force is weakening.