# PHY-2049C Lecture Notes - Lecture 1: Fermi Energy, Superposition Principle, Stationary Point

by OC1138882

PHY 2049C – Dr. Prosper/Y. Hori/W. Roberts

Week 1 – 01 (Tuesday Lecture)

Electric Charge, Force, and Field Pt. 1

• Electric Charge: Qualitative Facts

-There are two kinds of charge: positive (+) and negative (-)

-Charge is CONSERVED

-Lightning involves the flow of charge

-Charge can be transferred between objects

-Charge can be displaced

-Like charges repel, unlike charges attract

-Charge is quantized (it occurs in distinct units)

• Electric Charge: Polarization

-A metal is a lattice of positively charged ions immersed in a gas of electrons that

flows freely through the lattice.

-If a positively charged rod nears the metal, the freely floating gas of electrons nears

the positively charged rod. This creates a negative pole (near the positively charged

rod) and a positive pole (away from the positively charged rod) in the metal. This

process is known as POLARIZATION.

-An insulator is a lattice of electrically neutral atoms.

-When a charged rod nears an insulator, it polarizes the insulator (creates positive and

negative poles); HOWEVER, this is much less polarized than the metal.

• Electron Gas Pressure (“an aside” so I am not certain if this will be on the test)

-The pressure of the gas of free electrons in a metal is given by the approximate

formula:

P = (2/5) * ρ * Ef

Where ρ is the electron density (number of electrons per unit volume)

and Ef is the Fermi Energy: an energy characteristic of the material (would be given)

•Coulomb’s Law

-Coulomb’s Law states that: two stationary point charges q1 and q2 a distance r apart

exert a force on each other of magnitude:

o|F(1,2)| = k * q1 * q2 / r^2 = |F(2,1)|

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-Unit of charge is coulomb (C)

o1 C = 6.25 * 10^18 elementary charges

oTherefore, one elementary charge = e = 1.60 * 10^-19 C

oOne electron = -e; one proton = +e

-Follows superposition principle: the electric force on a point charge is the sum of the

Coulomb forces on it due to all other point charges.

•Coulomb Force Problems

1. Electric force due to a uniformly charged rod

“Find the electric force on a point charge Q1 at point P = (x, y) due to a thin uniformly

charged rod. Take the origin at the center of the rod. The total charge on the rod is Q2.”

Step 1: Define an element of charge

Imagine we are picking a point on the rod for this part. The charge on this rod is

evenly distributed! So, to find how much charge this point (dq) has: we need to find

the ratio of the length of this point (dl) to the length of the rod (L) and multiply it by

the total charge Q2. So..

dq = Q2*dl/L

Step 2: Write down the force of dq on Q1.

For this, we use Coulomb’s Law. r represents the distance between the dq and P

and k is a constant.

F2,1(vector) = [ (k * Q1 * dq) / (r^2) ] * r (unit vector)

Notice that we multiply by r hat (the unit vector of r) to make Coulomb’s Law

give the vector instead of just the magnitude. The r hat gives the vector its direction.

Now let’s substitute r in terms of x,y (the location of point P) and l (the position

of dq where –L/2 <= l <= L/2). r would be the hypotenuse of the right triangle with

legs of length x-l (horizontal) and y (vertical). Thus:

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r (vector) = (x-l) i + y j

[Note: i and j are unit vector representing the x axis direction and y axis direction,

respectfully. This is a notation you should know, though it is commonly written with

a ^ instead of dots.]

Using Pythagorean Theorem, we get:

|r| = [ (x-l)^2 + y^2 ]^0.5

Substituting into our vector equation we get:

F2,1 (vector) = [ (k * Q1 * dq) / ( (x-l)^2 + y^2) ] * r (unit vector)

Step 3: Write down the unit vector

The unit vector r hat is defined as: r(vector) / |r|

r (hat) = [ (x-l) / ( (x-l)^2 + y^2)^0.5 ]*i + [ y / ( (x-l)^2 + y^2)^0.5]*j

Step 4: Write down the net electric force on the point charge at P

Substituting r hat into our electric force equation and rewriting we get:

F2,1(vector) = k * Q1 * Q2 ( I_x * i + I_y j)

I_x = (1/L) {integral from –L/2 to +L/2} (x-l)dl / [ (x-l)^2 + y^2 ]^3/2

I_y = (y/L) {integral from –L/2 to +L/2} dl / [ (x-l)^2 + y^2 ]^3/2

Take a moment here to see what we did. The integral sign is here in order to sum up

the charge of all the points along the charged rod. The bounds for the integral was

where l was defined along: -L/2 to L/2.

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