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Coulomb's Law and Electrostatics: Intermolecular Interactions and Electric Fields, Study notes of Physical Chemistry

Hope College Physical Chemistry

An overview of coulomb's law, electrostatics, electric fields, and electrostatic potential. The lecture discusses various intermolecular interactions, including charge-charge, charge-dipole, dipole-dipole, and dipole-quadruple interactions. Coulomb's law is presented, and the calculation of electrostatic interaction between two charged particles is explained. The document also covers the concept of force, electric field, and potential, and their relationships.

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Lecture 35 Chapt 20 Coulomb’s Law (electrostatics)

Announce:

• Give exam distribution

Outline:

Coulomb’s Law

Electric Field

Electrostatic Potential

Intermolecular Interactions

Okay, for this week we are changing our focus to talk about various kinds of

intermolecular interactions.

Let’s look at the strongest interactions between molecules – electrostatics. (nuclear

forces are stronger, but that’s about it.)

Electrostatics includes charge-charge interactions, charge-dipole, dipole-dipole,

dipole-quadruple, etc. (often described as monopole/multipole interactions)

Coulomb’s law

Let’s start with Coulomb’s law which everyone remembers from physics

()

r

qq

ru BA

0

4

1

πε

=

u is the interaction energy (units J = V⋅C) (note W = J/s = V⋅C/s = V⋅A)

ε

0 is the permittivity of vacuum (units F/m = C/V⋅m = C2/J⋅m)

So, we can calculate the electrostatic interaction between two charged particles. In

practice this is a difficult calculation. Why? Because the interaction falls off very

slowly with distance ~ r

−

1

Text does an example of NaCl crystal. If you just look at one line of charges that

surround a central Na+:











+−+−+−= L

6

2

5

2

4

2

3

2

1

2

4

12

0a

e

u

πε

Partial preview of the text

Download Coulomb's Law and Electrostatics: Intermolecular Interactions and Electric Fields and more Study notes Physical Chemistry in PDF only on Docsity!

Lecture 35 Chapt 20 Coulomb’s Law (electrostatics)

Announce:

Give exam distribution

Outline :

Coulomb’s Law

Electric Field

Electrostatic Potential

Intermolecular Interactions

Okay, for this week we are changing our focus to talk about various kinds of

intermolecular interactions.

Let’s look at the strongest interactions between molecules – electrostatics. (nuclear

forces are stronger, but that’s about it.)

Electrostatics includes charge-charge interactions, charge-dipole, dipole-dipole,

dipole-quadruple, etc. (often described as monopole/multipole interactions)

Coulomb’s law

Let’s start with Coulomb’s law which everyone remembers from physics

( ) r

q q u r

A B

(^40)

u is the interaction energy (units J = V⋅C) (note W = J/s = V⋅C/s = V⋅A)

ε 0 is the permittivity of vacuum (units F/m = C/V⋅m = C

2

/J⋅m)

So, we can calculate the electrostatic interaction between two charged particles. In

practice this is a difficult calculation. Why? Because the interaction falls off very

slowly with distance ~ r

− 1

Text does an example of NaCl crystal. If you just look at one line of charges that

surround a central Na+:

= − + − + − + L

2

0 a

e u

we get something that converges VERY slowly. In this case there is a regular

arrangement of charges, so we can find a series approximation that helps us out

[defines the Madelung constant]. e.g.

a

N e U

A

2

This is great for cyrstalline solids, but for more general charge distributions (like

dissolved salts, amorphous materials, or proteins you have to try to just add them all

up.

Fortunately, in most media charges are shielded by the surroundings. This is

especially important in water. We can write Coulombs law the way it is really

defined

( ) r

q q u r

A B

= where ε = ε 0D where D is the dielectric constant

D ≡ 1 for vacuum

D = 1.00059 ≈ 1 for air

D ≈ 2 for hydrocarbons

D ≈ 1.4 often used for proteins

D = 33 MeOH

D = 78.54 for water

Book mentions Bjerrum length, which is a handy tool. Remember our golden tool

for connecting microscopic properties to macroscopic behavior? Partition function.

Energy always appears as E/kT or E/RT depending on units.

The Bjerrum length is just the distance two charged bodies are apart that

corresponds to U = RT.

Book points out that Bjerrum length is 80 times shorter ( i.e. a factor of D , 560 Å

versus 7.13 Å) for NaCl dissolved in water than for two bare Na

Cl

−

particles.

Their interaction is severely weakened by the presence of the water.

If we had many charges, we would just add up the u for each of them, until we had

accounted for everybody. However, there are easier ways, which we will work up

to now.

sphere and the flux would still be q/ ε 0. This also means that the total flux just depends on

the total charge contained, not how it is arranged.

q

Look familiar? Well, it is sort of Gauss’s law (more next time)

surface volume

q E ds dV

v v ρ

Potential

This leads us to our last important characterization of electrostatics, the electrostatic

potential, ESP. Just as the field was a uniform way to represent the force that might

affect a particle, the ESP is a way to represent the energy of a test charge. Formally,

it is the work needed to move the test charge between two points. To make things

consistent we always choose one test point that is at a region of zero ESP (usually

r = ∞). Remember that work is just the integral of force over distance, so in this

case it is the path integral of field from infinity to r. (Minus sign indicates that work

is done against the field rather than by the field.)

∞ = − ∞ = =− ⋅

r

test

r r E dl q

w v v

(units of ESP are work over charge, J/C)

Going back to our simple single point charge, we calculate the potential by moving

radially inward from infinity to some distance r. Again, E is always parallel to our

radial vector d r so we can dispense with the dot product.

Dr

q

r

dr

D

q dr Dr

q Edr

A

r A

r

0

2 0

2

∞ ∞ ∞

remember that the interaction energy between our test particle and the

charge(s) that generates the potential is just u(r) = q test ψ (r)

Note that we can get the field easily from the ESP. How might this be done?

if force is the negative gradient of the energy, then field is the negative

gradient of ESP.

E =−∇ ψ

v

Coulomb's Law and Electrostatics: Intermolecular Interactions and Electric Fields, Study notes of Physical Chemistry

Related documents

Partial preview of the text

Download Coulomb's Law and Electrostatics: Intermolecular Interactions and Electric Fields and more Study notes Physical Chemistry in PDF only on Docsity!

Lecture 35 Chapt 20 Coulomb’s Law (electrostatics)

Okay, for this week we are changing our focus to talk about various kinds of

intermolecular interactions.

Let’s look at the strongest interactions between molecules – electrostatics. (nuclear

forces are stronger, but that’s about it.)

Electrostatics includes charge-charge interactions, charge-dipole, dipole-dipole,

dipole-quadruple, etc. (often described as monopole/multipole interactions)

Coulomb’s law

Let’s start with Coulomb’s law which everyone remembers from physics

u is the interaction energy (units J = V⋅C) (note W = J/s = V⋅C/s = V⋅A)

ε 0 is the permittivity of vacuum (units F/m = C/V⋅m = C

/J⋅m)

So, we can calculate the electrostatic interaction between two charged particles. In

practice this is a difficult calculation. Why? Because the interaction falls off very

slowly with distance ~ r

Text does an example of NaCl crystal. If you just look at one line of charges that

surround a central Na+:

= − + − + − + L

we get something that converges VERY slowly. In this case there is a regular

arrangement of charges, so we can find a series approximation that helps us out

[defines the Madelung constant]. e.g.

This is great for cyrstalline solids, but for more general charge distributions (like

dissolved salts, amorphous materials, or proteins you have to try to just add them all

up.

Fortunately, in most media charges are shielded by the surroundings. This is

especially important in water. We can write Coulombs law the way it is really

defined

= where ε = ε 0D where D is the dielectric constant

D ≡ 1 for vacuum

D = 1.00059 ≈ 1 for air

D ≈ 2 for hydrocarbons

D ≈ 1.4 often used for proteins

D = 33 MeOH

D = 78.54 for water

Book mentions Bjerrum length, which is a handy tool. Remember our golden tool

for connecting microscopic properties to macroscopic behavior? Partition function.

Energy always appears as E/kT or E/RT depending on units.

The Bjerrum length is just the distance two charged bodies are apart that

corresponds to U = RT.

Book points out that Bjerrum length is 80 times shorter ( i.e. a factor of D , 560 Å

versus 7.13 Å) for NaCl dissolved in water than for two bare Na

Cl

particles.

Their interaction is severely weakened by the presence of the water.

If we had many charges, we would just add up the u for each of them, until we had

accounted for everybody. However, there are easier ways, which we will work up

to now.

sphere and the flux would still be q/ ε 0. This also means that the total flux just depends on

the total charge contained, not how it is arranged.

Look familiar? Well, it is sort of Gauss’s law (more next time)

v v ρ

Potential

This leads us to our last important characterization of electrostatics, the electrostatic

potential, ESP. Just as the field was a uniform way to represent the force that might

affect a particle, the ESP is a way to represent the energy of a test charge. Formally,

it is the work needed to move the test charge between two points. To make things

consistent we always choose one test point that is at a region of zero ESP (usually

r = ∞). Remember that work is just the integral of force over distance, so in this

case it is the path integral of field from infinity to r. (Minus sign indicates that work

is done against the field rather than by the field.)

(units of ESP are work over charge, J/C)

Going back to our simple single point charge, we calculate the potential by moving

radially inward from infinity to some distance r. Again, E is always parallel to our

radial vector d r so we can dispense with the dot product.

D

remember that the interaction energy between our test particle and the

charge(s) that generates the potential is just u(r) = q test ψ (r)

Note that we can get the field easily from the ESP. How might this be done?

if force is the negative gradient of the energy, then field is the negative

gradient of ESP.

E =−∇ ψ

Finally, the path integral formalism above gives us a clue that just like work against

gravity is conservative, work against an electric field is conservative. So, any closed path

will have an electrostatic work of zero.